Integration By Partial Fractions - Free Educational videos for Students in K-12

Integration By Partial Fractions - By The Organic Chemistry Tutor

Transcript


00:0-1	in this video , we're going to talk about how
00:02	to integrate rational functions used in partial fraction decomposition .
00:08	So the first thing we want to do is make
00:09	sure that this integral is completely factored on the bottom
00:14	. We have a difference of perfect squares . And
00:17	so we can factor that expression by taking the square
00:20	root of X squared which is X . and the
00:22	square root of four which is to one is going
00:25	to be positive to the other will be negative too
00:29	. So what we have on the bottom is two
00:32	distinct linear factors . Now you might be wondering what
00:36	is the difference between a linear factor and a quadratic
00:42	factor ? Linear factors include examples such as X three
00:53	XX Plus four , four X -5 . Those are
00:57	linear factors quadratic factors . They typically have an X
01:01	squared , so X squared X squared plus four .
01:06	That's a quadratic factor , X squared plus three .
01:10	Acts , that's another one or X squared plus two
01:13	X plus seven . Those are quadratic factors . Now
01:21	to use the technique of partial fraction decomposition , what
01:26	we're going to do is we're going to break up
01:27	this expression Into two smaller fractions . So here we
01:36	have X plus two on the bottom , we're gonna
01:39	have some constant Over X-plus two dx . And here
01:46	we have another linear factor in the bottom , so
01:48	we're going to have another content on top Over X
01:51	-2 DX . If we had a quadratic factor ,
01:55	let's say , like for instance , X squared plus
01:58	eight . Instead of putting a constant , like A
02:00	or B . Well put a linear factor A ,
02:04	X plus B . Or like bx plus C .
02:07	If we had a quadratic factor in the bottom .
02:10	Now , just to give you an idea of what
02:12	partial fraction decomposition is , let's say if we have
02:17	one plus four , I mean one of the four
02:19	plus 1/5 . If we wish to add these two
02:23	fractions , we would need to get common denominators .
02:27	If we multiply this fraction by 5/5 , it will
02:29	become 5/20 . If we multiply this by 4-4 ,
02:33	it will become 4 20 . So the sum of
02:39	these two fractions , one of the four plus one
02:42	of the five , It's 9/20 . Partial fraction decomposition
02:50	allows us to start With this fraction 9/20 and break
02:56	it down into these two simple fractions , one of
03:00	the four plus one of the five . So that's
03:03	a good example of partial fraction decomposition . We can
03:06	take a single fraction And break it up into two
03:09	smaller fractions . It could be two or more smaller
03:12	fractions , it could be three fractions , it could
03:14	be four . But that's the idea behind partial fraction
03:18	decomposition . We're taking a single fraction and break it
03:23	up into multiple fractions . So what we need to
03:28	do here is we need to determine the value of
03:30	A and B . And this problem . So for
03:34	now I'm going to get rid of the integral symbol
03:37	and the dx term . And we're going to set
03:41	this fraction equal to A Over X-plus two . Yeah
03:51	, plus B Over X -2 . If we can
03:56	calculate the value of A and B , then we
04:00	could use that to find the indefinite integral of those
04:04	two expressions . So we're going to do right now
04:07	, is we're gonna multiply every term . Bye .
04:11	The denominator on the left side . So that is
04:13	by X-plus two And X -2 . So when multiplying
04:21	this fraction by these two factors , This is going
04:25	to cancel and we're gonna be left for for one
04:28	on the left side of the equation . If we
04:31	take A over exports to and multiply by exports to
04:35	expand its too . The X-plus two factors will cancel
04:42	leaving behind A times X spanish too . Now let's
04:55	take this fraction multiplied by those two factors . So
04:59	X minus two will cancel . And we're gonna have
05:02	be times X plus two . Now , when you
05:12	have linear factors like X minus two , X plus
05:14	two , there's a shortcut technique that you could use
05:18	in order to calculate the value of A . And
05:20	B . And that is by plugging in certain X
05:23	values . If we were to plug in X equals
05:26	two , This term becomes zero so we're gonna have
05:31	one is equal to eight times two minus two to
05:34	minus 20 So this becomes zero plus B . And
05:38	this is going to be two plus two , which
05:40	is four . So what we have is one is
05:43	equal to for me And then dividing both sides by
05:46	four . We get that be Is equal to 1/4
05:59	. Now another value that we can plug in Is
06:02	-2 . If we plug a negative too negative two
06:05	plus two becomes zero . So this whole thing becomes
06:08	zero . And so what we're gonna have is one
06:11	is equal to a . Times -2 managed to and
06:17	negative two minus two is negative four , Dividing both
06:21	sides by -4 . We'll get that a . Is
06:24	-1/4 . So now we have the values of A
06:29	. And B . So going back to this expression
06:34	let's replace A . With negative 1/4 and then let's
06:42	replace B . With positive 1/4 . Now what we're
06:59	gonna do is we're going to take these constants and
07:02	move them to the front . So this is gonna
07:09	be negative 1/4 integral one over X plus two dx
07:18	And then plus positive one of the four integral one
07:21	over X minus GDX . Now let's talk about how
07:26	we can integrate rational functions . The anti derivative of
07:31	one of her ex is Ellen X . Now we
07:37	need to use an absolute value symbol because we can
07:41	have a negative value inside a natural log function .
07:45	It's not gonna work , it's gonna be undefined .
07:50	So let's say if we want to integrate one over
07:53	X plus five it's going to be the natural log
07:57	of X plus five . If we have a constant
08:01	on top of that let's say four over extra seven
08:06	. This is going to be it's L . N
08:09	. X plus seven . But times for now let's
08:14	see if we have the anti derivative of 1/2 X
08:18	plus seven . This is going to be Ellen two
08:23	x plus seven but we need to take into account
08:27	the two in front of the attacks , The derivative
08:30	of X is one . So when you divide this
08:32	by one , the answer doesn't change . The derivative
08:35	of two X . is too . So we need
08:38	to divide this by two . It turns out this
08:40	is gonna be one half Ln two X plus seven
08:44	. Likewise , if we want to find the anti
08:46	derivative of 1/3 X plus eight , this is going
08:52	to be Ellen do we act plus eight ? But
08:55	then Divided by the derivative of three X , which
08:58	is one of the three . And of course all
09:01	of these have plus see the constant of integration .
09:06	Now let's see if we have seven Divided by five
09:10	x plus nine . So this one we have to
09:15	be more careful . This is gonna be Ln five
09:18	x plus nine . But we need to divide this
09:21	by The derivative of five x . Which is going
09:25	to be five . So it's one of the five
09:29	if you're dividing it And then we have to multiply
09:31	by seven . So it's time seven times 1/5 Ln
09:36	five X plus nine . So you're going to be
09:40	using this a lot when dealing with partial fraction integration
09:48	. So I want to just give you some ideas
09:50	on how to integrate rational functions into natural algorithmic functions
09:58	. So the anti derivative of one over X plus
10:00	two , that's simply going to be Ln exports to
10:06	And the anti derivative of X -2 is going to
10:09	be Ln I mean the inside derivative of one of
10:13	our x minus two , S L n x minus
10:16	two . Now let's add the constant of integration .
10:23	Now , what I'm gonna do is I'm going to
10:24	factor out one of the four and I'm going to
10:27	write the positive term first in the negative term .
10:30	Second , It's a factory that one of the four
10:34	. We're going to get Ellen x minus two minus
10:41	Ln x plus G . Now a property of logs
10:49	is that we can convert to logs into a single
10:53	log . For instance , let's say we have L
10:56	N A minus Ln B . We can write this
10:59	as Ellen A over B . Or if we have
11:03	L N A plus L N B , we can
11:06	write this as a single log Ln A times B
11:10	. But for this problem we have a minus sign
11:13	. So we're going to use that property of natural
11:16	logs , which applies for all logs . So we
11:21	could rewrite this answer as one of the four times
11:27	Ellen absolute value . This is gonna go on top
11:30	since it's positive , this is going to go on
11:37	the bottom since it has a negative sign in front
11:40	of it and then plus C . So this is
11:48	the final answer for this problem . So that is
11:54	the anti derivative of one over x squared minus four
11:58	. So that's how we can get that answer .
12:00	Use an integration by partial fractions . Now , let's
12:04	work on a similar example for the sake of practice
12:07	, so feel free to pause the video . If
12:09	you want to try this example , We have the
12:11	integral of X -4 over X squared plus two X
12:16	minus 15 . So go ahead and try this problem
12:20	. The first thing we need to do is we
12:22	need to factor this expression completely , particularly the denominator
12:27	and what we have is a try no meal Where
12:29	the leading coefficient is one . So we need to
12:32	find two numbers That multiply to the constant negative 15
12:37	but add to the middle coefficient to So this is
12:41	going to be positive five and negative three . Five
12:46	plus negative three adds up to two but multiplies to
12:49	negative 15 . So it's a factor . This expression
12:53	, It's going to be X-plus five times X minus
12:58	tree . So here we have two distinct linear factors
13:04	. So we can write this as the integral of
13:07	a Over X-plus five dx plus the integral of B
13:15	over ex monastery dx . So like last time we
13:19	need to calculate the value of A and B .
13:22	And then we could find the answer . So let's
13:29	set this fraction equal to the two fractions that we
13:34	have on the right , just without the integral symbol
13:38	. So this is going to be a Over X-plus
13:41	five and then plus B over X monastery . And
13:49	just like before we're gonna multiply everything . Bye .
13:53	And those two factors . So multiplying this fraction by
14:03	X-plus five times x monastery . And these will cancel
14:08	. And we're going to get X -4 on the
14:10	left side of the equation . This fraction time is
14:17	those two factors X-plus five will cancel . And we're
14:20	gonna have a times X minus street left over .
14:30	And then when we multiply be over ex monastery by
14:32	these two factors , the x minus three factor will
14:37	cancel . Leaving behind B times X plus five .
14:46	Select before we're going to plug in X values to
14:49	calculate the value of A . And B . So
14:51	let's begin by plugging in three . If we plug
14:55	in X equal street This term goes to zero .
14:59	So we're going to have 3 -4 is equal to
15:03	zero Plus A . B . Times three Plus 5
15:08	. Now 3 -4 is negative one , three plus
15:11	5 is eight , Dividing both sides by eight .
15:16	We're gonna get that B is equal to negative 1/8
15:24	. So now let's go ahead and calculate A .
15:29	What X . Value ? Do we need to plug
15:31	in to get A . To find the X .
15:36	Value ? You could set X plus five equal to
15:38	zero and soft rex . If you do you get
15:42	X . Is equal to negative five . So that's
15:46	we're gonna plug in we're gonna plug in -5 .
15:51	So this becomes negative 5 -4 is equal to a
15:56	. Times -5 -3 . And then negative 5-plus 5
16:02	will become zero . But you can write it out
16:03	to show your work if you want to -5 -4
16:06	is negative nine , -5 -3 is negative eight and
16:11	then negative five plus 50 dividing both sides by negative
16:16	. Ain't we get that ? A . Is equal
16:20	to positive 9/8 . The two negative signs will cancel
16:26	. So now that we have the values of A
16:28	. And B . We can go ahead and plug
16:30	it in into the original expression . So this becomes
16:36	9/8 and this is gonna be -1/8 . So we
16:54	can rewrite this size 9/8 times the integral of one
17:00	over X plus five dx And then -1/8 Times The
17:05	integral of one over X minus street dx . The
17:12	anti derivative of one over X plus five is going
17:15	to be the natural log of X plus five .
17:21	And then the anti derivative of one of her ex
17:22	monastery is going to be Ln ex monastery and then
17:28	plus C . Since the constant in front of Ellen
17:34	are different , it might be good to leave it
17:37	in this form . Now you can still combine it
17:41	into its into a single log expression if you want
17:44	to , If you wish to do that . What
17:47	I would do is move the nine here . A
17:50	property of logs allows you to move the coefficient to
17:53	the export position . For instance to L N X
17:56	is equal to L N X square . You can
17:59	move it to to the exploded position of X .
18:07	So first I would rewrite it as 1/8 Ellen and
18:12	this would be X-plus 5 to the knife power but
18:17	still within An absolute value simple and then -1/8 Ellen
18:23	ex monastery plus C . And then we can factor
18:27	out 1/8 at this point and then write it as
18:32	a single lot of expression . So it's gonna be
18:34	one of her eight Ln absolute value X-plus five race
18:40	at a knife power over X minus stream . And
18:44	then plus C . So you can invite it like
18:47	this as a single log if you wish . So
18:53	that's it for this problem . So let's try a
18:57	slightly harder problem . So in this problem we have
19:01	the linear factor X -1 . But we also have
19:04	a repeated Linear Factor X -2 . So how can
19:09	we set up this problem ? Well , what we're
19:13	gonna do is we're going to write three fractions instead
19:16	of two . We're going to have the integral of
19:21	a Over X -1 and then we're going to have
19:30	the integral of B over the other , linear factor
19:35	x minus two . And since it's x minus two
19:40	squared , we're going to have another fraction . This
19:43	one's going to B C Over X -2 Squared .
19:50	So that's what you need to do . If you
19:52	have a repeated linear factors . For instance , let's
19:56	say if we have one over xmas history To the
20:01	3rd power to break it up using partial fractions ,
20:06	this would be a over x minus stream , plus
20:10	a B Over X -3 Squared and then plus C
20:15	over x minus street cube . So that's how we
20:18	would set up the problem . If this was a
20:21	race to the 3rd power , Let's go ahead and
20:26	work on this one . So let's rewrite it without
20:33	the integral symbols . So we're gonna multiply every fraction
20:56	By X -1 times x -2 squared . So when
21:03	multiplying these two , these will completely canceled . Leaving
21:08	behind X on the left side of the equation when
21:14	multiplying these two , X -1 will cancel and we'll
21:22	be left with a Times X -2 Squared . Next
21:38	when multiplying needs to only one Of the two X
21:44	-2 factors will cancel . So we're gonna have to
21:47	be Times X -1 times x -2 . And finally
22:01	the X -2 squared terms will cancel . And we'll
22:04	be left with c . times the factor X -1
22:09	. Now what we're gonna do is we're gonna plug
22:11	in some numbers . Let's focus on The X -2
22:17	Factor . So we're going to plug in is we're
22:20	gonna plug in X equals two . This is going
22:22	to be too and then this is going to be
22:25	a 2 -2 is zero and then be two minus
22:30	one is one , 2 -2 is cyril and then
22:33	plus C 2 -1 is one . So we get
22:37	two is equal to see because those they go to
22:40	zero . So let's write that here , C is
22:44	equal to two . Now since we have some X
22:51	-1 factors We're going to plug in x equals one
22:57	. So this is going to be one A And
23:00	then 1 -2 squared plus a . B 1 -1
23:05	is zero And for C It's 1 -1 which is
23:09	also zero . 1 -2 Squared . That's negative one
23:15	squared Which is a positive one . So basically a
23:20	. Is equal to one . Now to get B
23:28	. We need to plug in some other number that's
23:31	not one or two . So let's pick the next
23:34	best number which is three . This is going to
23:37	be three . A three minus two is 11 squared
23:40	is one And then 3 -1 is 2 3 -2
23:45	is one And then 3 -1 is two . Now
23:50	we know A . Is one so that becomes one
23:54	and C . Is to so two times two is
23:56	four And 1-plus 4 is five . That is adding
24:05	those two subtracting both sides by five . We're going
24:08	to have three minus five which is negative two .
24:11	And then if we divide both sides by two we'll
24:14	get that be Is equal to -1 . So let's
24:22	replace a with one unless you place be with negative
24:34	one and then see with two The anti derivative of
24:50	one over X -1 is going to be Ln x
24:54	minus one an absolute value . And the next this
24:58	is gonna be negative . Ellen x minus two for
25:06	this one . We could use use substitution . So
25:10	I'm going to rewrite the expression as two times the
25:12	integral of one Over X -2 Squared . So we're
25:18	going to make you equal to X -2 D'you is
25:22	going to equal dx . So this becomes two times
25:27	the integral of one over U squared . Do you
25:32	moving to you variable to the top ? This becomes
25:34	too integral U . To the negative too . Now
25:39	we can integrate using the powerful so we're going to
25:42	add one to the negative two exponents . So it
25:44	becomes U . To the negative one . And then
25:47	we're gonna divided by negative one bringing you back to
25:51	the bottom , this becomes negative to over you And
25:54	then replacing you with X -2 , It becomes -2
25:58	over X -2 . So this is the answer .
26:11	Now if we want to we can write the to
26:15	log expressions into a single log . So we can
26:18	write as Ellen X -1 over X -2 And then
26:24	-2 Over X -2 Plus C . So we can
26:30	leave the answer like this if we wish . So
26:33	that's it for problem number three go ahead and try
26:38	this problem . Find the anti derivative of X square
26:41	plus nine over x squared minus one times x squared
26:45	plus four . So as always you want to make
26:49	sure that the denominator the fraction is completely factored .
26:55	We can't factor x squared plus four but we can
26:58	factor x squared minus one . Since we have a
27:01	difference of perfect squares . The square root of x
27:04	squared is X . The square root of one is
27:07	one . So we're gonna have X plus one and
27:09	x minus one . So what we have here is
27:18	to linear factors and the quadratic factor . So we
27:24	could set this up as a Over X-plus one and
27:33	then plus B Over X -1 . Now , for
27:41	the quadratic factor x square plus four on top ,
27:44	we're going to write C . X plus team instead
27:47	of C or D by itself . So that's how
27:56	we could set up the partial fraction decomposition . Now
28:05	on top , we can write this as the integral
28:08	of A over X plus one dx and then plus
28:13	the integral of B Over X -1 , the X
28:18	. And then plus the integral of C . X
28:22	. Plaza de over X squared plus four . Now
28:29	let's multiply both sides of the equation by this denominator
28:33	here . So I'm just going to write this on
28:38	this side . So we have X plus one ,
28:41	X minus one and X squared plus four . So
28:46	these will completely cancel . Leave behind X square plus
28:49	nine on the left side . And then this fraction
28:56	times all of that . The X plus one fact
29:00	the X plus one factors will cancel . Leave behind
29:03	A Times X -1 times X squared plus four .
29:09	Next we're going to play this fraction by that X
29:14	-1 will cancel . So we're going to have be
29:19	Times X-plus one times X squared plus four . It's
29:26	important to be very careful with every step because if
29:29	you make one mistake , this is a long problem
29:32	. The whole problem is is going down the drain
29:35	so it's going to take your time working with these
29:39	problems . So now multiplying this by what we have
29:45	here , X squared plus four will cancel . Leaving
29:49	behind . See explicit E Times X-plus one Times X
29:54	-1 . Now the term X squared plus four will
29:59	always be greater than zero . So there's no X
30:03	value that we can plug in to make that disappear
30:06	. Now we do have some linear factors X -1
30:09	and X plus one . So we can plug in
30:11	X equals positive one and X equals negative one .
30:14	To turn those linear factors into zero . So let's
30:18	start by plugging X equals positive one first . So
30:24	we're gonna have one square plus one . I mean
30:27	one square plus nine rather and then A times 1
30:33	-1 is going to be zero . So this Entire
30:36	term becomes zero and then plus B . one Plus
30:40	1 is two . One square plus four . That's
30:43	going to be five . And here we have an
30:47	X -1 factor . So when we plug in one
30:49	That entire term here is gonna be zero . So
30:53	one plus 9 is 10 , two times 5 is
30:56	10 . So we have 10 is equal to 10
30:58	B . The volume both sides by 10 . We
31:02	get that B is equal to one . So let's
31:07	go ahead and replace be with one . So now
31:17	that we use X equals one . Let's try plugging
31:21	in x equals negative one . -1 Squared is still
31:35	one . So we're going to have one plus nine
31:37	which is 10 on the left side and then it's
31:41	gonna be a 1 -1 is negative two negative one
31:45	squared plus four . That's going to be positive one
31:48	plus four which is five . When we plug a
31:52	negative one into X . Plus one , that's going
31:54	to be zero and this will be zero sua So
31:58	we get 10 is equal to negative 10 . A
32:02	dividing both sides by negative 10 . We can see
32:05	that A . Is negative one . Now what's the
32:19	next X . Value that you think we should plug
32:21	in ? I would recommend plug it in X .
32:26	equals zero . We have the valleys for A .
32:30	And B . We have to focus on Cnd if
32:33	we plug in zero C disappears which means we only
32:37	have the leftover which means we could solve for D
32:41	. So on the left we're gonna have zero squared
32:43	plus nine . This is going to equal A .
32:46	We know A . Is negative one And then 0
32:51	-1 is -1 . Zero square plus four . That's
32:55	four . Be as positive one . zero Plus 1
33:00	is one . And cyril squared plus four . That's
33:03	going to be four . Again . See explicit E
33:05	. This is going to be see time zero plus
33:11	the D . Zero plus one is 10 minus one
33:14	is negative one . So this is nine negative one
33:18	times negative one times voice four . And then we
33:20	have another four . See time zero plus D .
33:24	Is just A . D . And then we have
33:27	one times negative one . So this is nine is
33:30	equal to eight -D . Subtracting both sides by eight
33:34	. We have 9 -8 which is -1 . And
33:38	that's equal actually that's positive one and that's equal to
33:42	negative D . So D . Is equal to negative
33:45	one . So let's put negative one here for D
34:00	. Now we need to calculate C . To do
34:06	that . Let's pick the next best X . Value
34:09	. And uh Let's go with x equals two .
34:19	Since we know A . B . And C .
34:21	I mean we know a BND . So now we
34:23	can just find the missing constancy . So plugging into
34:32	for X . This is going to be two squared
34:34	plus nine . A . is negative one . Two
34:38	minus one is 12 squared plus four . That's going
34:41	to be eight . B . Is 12 plus one
34:45	. History and two squared plus four is eight .
34:47	Again . See explicit E . We're looking for C
34:52	. We have X which is two . So this
34:54	is going to be C . Times two or to
34:56	see D . is -1 . And then we have
35:02	two plus one which is three 2 -1 which is
35:05	one . Two squared is 44 plus nine is 13
35:10	. And then we have negative eight plus 24 .
35:14	And then if we distribute three it's a two C
35:16	minus one That becomes six c . Ministry negative eight
35:25	plus 24 That's gonna be 16 . And then we
35:31	have 16 -3 which is 13 . Subtracting 13 from
35:38	both sides 13 -13 0 . And so c .
35:42	is equal to zero . So this becomes zero x
35:48	. Which basically it disappears . So now that we
35:53	have the values of A . B . C .
35:54	And D . We can go ahead and determine the
35:59	integral for this original expression . So the integral of
36:06	negative one over X plus one . That's gonna be
36:08	negative Ln Absolute value expressed one and this is going
36:13	to be plus The Natural Log Absolute Value X -1
36:20	. Now this part here there's a formula that can
36:24	help us to integrate it . If you wish to
36:27	find the integral of dx over X squared plus a
36:33	squared . This is going to be won over A
36:37	. Are tangent X over A plus C . So
36:44	in this example we can see that A square is
36:51	four . So therefore a . Is too . Now
36:54	we do have a negative one . So we need
36:56	to put a negative here is gonna be negative one
36:58	over A . Or negative 1/2 are tangent X over
37:03	two Plus C . So that's gonna be the integral
37:07	of that expression . Now , if for some reason
37:13	you're not allowed to use that formula and you have
37:15	to show your work , here's what you can do
37:22	. So I'm going to move the negative sign to
37:23	the front . So let's start with this expression we're
37:27	going to use trig substitution . So we're gonna make
37:32	X . Or rather we're gonna make X squared equal
37:35	to four tan squared . So X is going to
37:40	be two tangent data . Dx is going to be
37:44	two times the derivative of tangent which is seeking squared
37:48	theta and then time's D . Data . So this
37:53	is going to be we're going to replace the dX
37:56	with to seek in squared theta . D . Theta
38:02	On the bottom . We have four Tan Square data
38:06	and then plus four . So I replaced the X
38:09	squared dx . Now what we're gonna do now is
38:23	we're gonna factor the four . So we're gonna be
38:27	left with Tangent squared plus one To over four reduces
38:35	to 1/2 . So I'm going to move the one
38:37	half to the front . So we have negative one
38:39	half integral , seek and squared D . Theta one
38:44	plus tan squared is equal to seek and squared .
38:49	So we could cancel seeking squared . So this becomes
38:52	negative 1/2 integral D . Theta which is the integral
38:57	of the fatah is simply fail to . Now we
39:00	need to find out what that is equal to .
39:03	So in this equation I'm going to divide by two
39:05	so we get X over two is equal to tangent
39:07	data . If we take the arc tangent of both
39:10	sides of this equation , we'll get our tangent X
39:15	over two is equal to our tangent of tangent of
39:19	theta . They are tan and tan cancel given us
39:22	data . So data is are tangent X over two
39:30	. And then we could put plus C . So
39:32	that's how you can show your work converting this expression
39:38	into an arc tangent expression . But if you know
39:40	the formula , it will save you time at least
39:43	you know what to do . So what I recommend
39:46	doing if you have to show your work is replace
39:48	the X squared term With four tangent squared . The
39:52	reason why I chose 400 square is because of the
39:54	four here . So you can factor out of four
39:57	. Get tanned square plus one and convert that into
40:00	second squared . So if this was X squared plus
40:03	nine , I would make X squared nine tangent squared
40:14	. So now let's go ahead and finish this problem
40:17	. What we can do is combine this into a
40:18	single log expression , so this is gonna be the
40:21	natural log , we're going to put the positive one
40:23	on top , so it's X minus one over X
40:27	plus one and then minus one half arc tangent X
40:34	over two Plus C . So that's the answer for
40:39	this problem .

Summarizer

DESCRIPTION:

This calculus video tutorial provides a basic introduction into integrating rational functions using the partial fraction decomposition method. Partial fraction decomposition is the process of breaking a single complex fraction into multiple simpler fractions. The integrals of many rational functions lead to a natural log function with absolute value expressions. This video explains what to do when you have repeated linear factors and quadratic factors. This tutorial contains many examples and practice problems on integration by partial fractions.

OVERVIEW:

Integration By Partial Fractions is a free educational video by The Organic Chemistry Tutor.

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