# Integration

• Apr 27th 2010, 05:07 PM
Mathbo
Integration
Find the following integrals:

a) integral from -1 to 0 of 1/(sqrt of 1-x) dx
b) indefinite integral of x(ln(x))^2 dx
c) indefinite integral of (x+7)/((x-3)(x+2)) dx

My final exam is almost here. I need help!
This is a practice question and for part a, I got -2+2sqrt2
Part b: (ln(x))^2 * x^2/2 - lnx * x^2/2 - x^2/4 + C
Part C: ln of absolute value of (x-3)^2/(x+2) +C
• Apr 27th 2010, 05:24 PM
Quote:

Originally Posted by Mathbo
Find the following integrals:

a) integral from -1 to 0 of 1/(sqrt of 1-x) dx
b) indefinite integral of x(ln(x))^2 dx
c) indefinite integral of (x+7)/((x-3)(x+2)) dx

My final exam is almost here. I need help!
This is a practice question and for part a, I got -2+2sqrt2
Part b: (ln(x))^2 * x^2/2 - lnx * x^2/2 - x^2/4 + C
Part C: ln of absolute value of (x-3)^2/(x+2) +C

Yes all fine.
• Apr 27th 2010, 05:34 PM
Mathbo
Quote:

Yes all fine.

Seriously?
What method am I supposed to use for part C?
• Apr 27th 2010, 05:52 PM
Quote:

Originally Posted by Mathbo
Seriously?
What method am I supposed to use for part C?

In your post you said you solved it so whatever method you used is fine. Probably partial fractions or something like that. It's 3am so I can't give it tooo much thought...
• Apr 27th 2010, 06:35 PM
Mathbo
Quote:

In your post you said you solved it so whatever method you used is fine. Probably partial fractions or something like that. It's 3am so I can't give it tooo much thought...

Did you do the actual integration and actually get the same answers as I did? I'm worried that I have problems with integration. And my final is tomorrow!
Thanks!
• Apr 27th 2010, 07:02 PM
harish21
Quote:

Originally Posted by Mathbo
Did you do the actual integration and actually get the same answers as I did? I'm worried that I have problems with integration. And my final is tomorrow!
Thanks!

use partial fractions to get:

$\displaystyle \int \frac{x+7}{x-3)(x+2)} dx = \int [\frac{2}{x-3} -\frac{1}{x+2}] dx$

and integrate
• Apr 28th 2010, 01:19 AM
rooshidavid
Well i too solve it. Your answer is right. But the calculations to shown on the post is so difficult so i suggest you to just be relax that you are right in your calculation. Also i would suggest you to don't delete this post because it could be helpful you to ask and take solutions in your any other difficulties.
• Apr 28th 2010, 02:10 AM
Quote:

Originally Posted by Mathbo
Did you do the actual integration and actually get the same answers as I did? I'm worried that I have problems with integration. And my final is tomorrow!
Thanks!

I did it with Maple. I'd say you'll be fine.
• Apr 28th 2010, 05:51 AM
edison
By partial integration

∫{(x+7)/(x-3)(x+2)}dx=∫{A/(x-3)+B/(x+2)}dx

A(x+2)+B(x-3)=x+7

Let x=3

5A=10

A=10/5
A=2

Let x=-2

-5B=5

B=-1

∫{(x+7)/(x-3)(x+2)}dx=∫{2/(x-3)-1/(x+2)}dx

=∫{2/(x-3)dx- ∫1/(x+2)}dx

=2ln(|x-3|)-ln(|x+2|)+C