Partial fractions for integration

**celtic1234** · September 10th 2011, 03:16 AM

Hi
Can you tell me if my solution is correct?
thanks
John

Evaluate using the method of partial fractions:

$\frac {10.dx}{(2x-6)(x+5)}$

This gives

$10=A.(x+5)+B.(2x-6)$

Making
$x=-5$

$\frac{1}{-6}=B$

$x=3$

$\frac{5}{4}=A$

This results in an separate integrations of:

$5\int \frac{1}{4.(2x-6)}- \int \frac{1}{6(x+5)}$

This solves to give;

$\frac{5}{4} ln(2x-6)-\frac{1}{6}ln(x+5)+C$

**HallsofIvy** · September 10th 2011, 03:28 AM

What you are doing is correct but your write it in such a careless way that you give the wrong answer!

Originally Posted by celtic1234

Hi
Can you tell me if my solution is correct?
thanks
John

Evaluate using the method of partial fractions:

$\frac {dx}{(x^2-1)}$

Below the line factors are

$\(x-1), (x+1)$

$\int \frac {1}{(x^2-1)}.dx=\frac{A}{(x-1)}+\frac{B}{(x+1)}$

The integral sign should not be here. You mean simply
$\frac{1}{x^2- 1}= \frac{A}{x- 1}+ \frac{B}{x+1}$

This gives

$1=A.(x+1)+B.(x-1)$

Making
$x=-1$

$\frac{1}{-2}=B$

$x=1$

$\frac{1}{2}=A$

This results in an separate integrations of:

$\int \frac{1}{2(x-1)}- \int \frac{1}{2(x+1)}$

This solves to give;

$\frac{1}{2} ln(x-1)-ln(x+1)+C$

No. The way you have written it, the "1/2" only multiplies the first ln and you want it with both:
$\frac{1}{2} ln(x-1)- \frac{1}{2}ln(x+1)+ C$
which is, of course, also equal to
$\frac{1}{2}ln(\frac{x-1}{x+1})+ C= ln\left(\sqrt{\frac{x-1}{x+1}}\right)+ C$

**celtic1234** · September 10th 2011, 03:34 AM

sorry i completely editing the latex and the problem?i have been studying for last 7 hours- i need to take a break....tired-somehow started to post a problem that i had the solution for? apologies

**Sudharaka** · September 10th 2011, 03:39 AM

Originally Posted by celtic1234

Hi
Can you tell me if my solution is correct?
thanks
John

Evaluate using the method of partial fractions:

$\frac {10.dx}{(2x-6)(x+5)}$

This gives

$10=A.(x+5)+B.(2x-6)$

Making
$x=-5$

$\boxed{\frac{1}{-6}=B}$ <---------Incorrect. The correct answer is, B=-5/8

$x=3$

$\frac{5}{4}=A$

This results in an separate integrations of:

$5\int \frac{1}{4.(2x-6)}- \int \frac{1}{6(x+5)}$

This solves to give;

$\frac{5}{4} ln(2x-6)-\frac{1}{6}ln(x+5)+C$

Please do not alter the original post after a solution is posted by someone. It causes confusion, specially for somebody seeing the thread after several posts are made. Always create new threads for new questions.

**celtic1234** · September 10th 2011, 03:55 AM

thanks I see my mistake
My problem is i am trying to post using Latex and need to edit the post after it's posted-by the time i edit it correctly somebody has replied....

I presume the correct answer is;

$\frac{5}{4}ln(2x-6)-\frac{5}{8}ln(x+5)$

**Sudharaka** · September 10th 2011, 04:17 AM

Originally Posted by celtic1234

thanks I see my mistake
My problem is i am trying to post using Latex and need to edit the post after it's posted-by the time i edit it correctly somebody has replied....

I presume the correct answer is;

$\frac{5}{4}ln(2x-6)-\frac{5}{8}ln(x+5)$

In obtaining the answer you have taken (which I did not notice in my previous post), $\int\frac{dx}{2x-6}=\ln(2x-6)$ which is incorrect.

I hope you know that the integration of the function $\frac{1}{x+a}$ gives,

$\int\frac{dx}{x+a}=\ln|x+a|+C$

Hence, $\int\frac{dx}{2x-6}=\frac{1}{2}\int\frac{dx}{x-3}=\frac{1}{2}\ln|x+3|$

Also you haven't put the modulus sign when taking the integrations.

$\frac{5}{4}\int \frac{dx}{2x-6}- \frac{5}{8}\int \frac{dx}{x+5}$

$\frac{5}{8}\int \frac{dx}{x-3}- \frac{5}{8}\int \frac{dx}{x+5}$

$\frac{5}{8}\left(\ln|x-3|-\ln|x+5|\right)+C$

$\frac{5}{8}\ln\left|\frac{x-3}{x+5}\right|+C$

**celtic1234** · September 10th 2011, 10:22 PM

thanks Sudharaka-see my mistakes-too tired yesterday.....

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