integrating a problem

**L_U_K_E** · February 24th 2008, 05:18 PM

int(2x/(x-3)^2) dx ..??? not sure how to do this.

**Jhevon** · February 24th 2008, 05:20 PM

Originally Posted by L_U_K_E

int(2x/(x-3)^2) dx ..??? not sure how to do this.

substitution. $u = x - 3$

**L_U_K_E** · February 24th 2008, 05:23 PM

Originally Posted by Jhevon

substitution. $u = x - 3$

i dont think that would help.. then du=dx and it would be int(2x/u)du and the x is still in there.

**Jhevon** · February 24th 2008, 05:25 PM

Originally Posted by L_U_K_E

i dont think that would help.. then du=dx and it would be int(2x/u)du and the x is still in there.

if $u = x - 3$

then $x = u + 3$

$\Rightarrow 2x = 2u + 6$

use that expression to replace 2x and you're good

**L_U_K_E** · February 24th 2008, 05:35 PM

Originally Posted by Jhevon

if $u = x - 3$

then $x = u + 3$

$\Rightarrow 2x = 2u + 6$

use that expression to replace 2x and you're good

ok ya your right. makin sense but i get down to it and the real question was integral from 0 to 2 of (2x/(x-3)^2) so u=x-3 du=dx 2x=2u+6 and i get..... int from 0 to 2 ((2u +6)/u du which is int from 0 to 2 of (2u/u) du + int from 0 to 2 of (6/u) du and then that is 4 - 6 ln |x - 3 | from 0 to 2 and i get 4 - 6 ln3 ... but the back of the book says 4-ln9 which is a different answer... so did i go wrong somehwerE?

**Jhevon** · February 24th 2008, 05:39 PM

Originally Posted by L_U_K_E

ok ya your right. makin sense but i get down to it and the real question was integral from 0 to 2 of (2x/(x-3)^2) so u=x-3 du=dx 2x=2u+6 and i get..... int from 0 to 2 ((2u +6)/u du which is int from 0 to 2 of (2u/u) du + int from 0 to 2 of (6/u) du and then that is 4 - 6 ln |x - 3 | from 0 to 2 and i get 4 - 6 ln3 ... but the back of the book says 4-ln9 which is a different answer... so did i go wrong somehwerE?

so there are two ways to go here. it is mainly a matter of personal preference. i am the kind of guy that, unless it causes way too much trouble, leaves the limits in tact in terms of x. when i integrate in terms of u i just rewrite the answer in terms of x and use the old limits

the other option is to change the limits so that you can use them with u. in that case, you use your substitution equation.

we have u = x - 3

so when x = 0, u = -3

when x = 2, u = 1

so your limits in u is from -3 up to 1, not 0 up to 2

**L_U_K_E** · February 24th 2008, 05:42 PM

Originally Posted by Jhevon

so there are two ways to go here. it is mainly a matter of personal preference. i am the kind of guy that, unless it causes way too much trouble, leaves the limits in tact in terms of x. when i integrate in terms of u i just rewrite the answer in terms of x and use the old limits

the other option is to change the limits so that you can use them with u. in that case, you use your substitution equation.

we have u = x - 3

so when x = 0, u = -3

when x = 2, u = 1

so your limits in u is from -3 up to 1, not 0 up to 2

ok ya. i think then that comes out to be the same answer though.. just 4-6ln3... but the answer is supppose to be 4 - ln9 ... and i dont get why considering there is an obvious 6 in the part where its int(6/u) du ... so idk wat went wrong.

**Jhevon** · February 24th 2008, 05:47 PM

$\int \frac {2x}{(x - 3)^2}~dx$

Let $u = x - 3 \implies x = u + 3$

$\Rightarrow du = dx$

So our integral becomes:

$\int \frac {2u + 6}{u^2}~du$

$= \int \left( \frac 2u + 6u^{-2} \right)~du$

$= 2 \ln |u| - \frac 6u + C$

the problem, i believe, is you divided by u rather than u^2

**L_U_K_E** · February 24th 2008, 05:48 PM

Originally Posted by Jhevon

$\int \frac {2x}{(x - 3)^2}~dx$

Let $u = x - 3 \implies x = u + 3$

$\Rightarrow du = dx$

So our integral becomes:

$\int \frac {2u + 6}{u^2}~du$

$= \int \left( \frac 2u + 6u^{-2} \right)~du$

$= 2 \ln |u| - \frac 6u + C$

the problem, i believe, is you divided by u rather than u^2

wow omg. im an idiot. i forgot the u^2 in the bottom. thank you so much.

**Jhevon** · February 24th 2008, 05:49 PM

Originally Posted by L_U_K_E

wow omg. im an idiot. i forgot the u^2 in the bottom. thank you so much.

you're not an idiot. we all miss things like that sometimes. just check around for my posts!

the important thing is, when you are practicing, you notice that you make such mistakes, so you look out for them on the test. don't beat yourself up

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