Partial fractions for integration

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September 10th 2011, 03:16 AM
celtic1234

Partial fractions for integration

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;

$\frac{5}{4} ln(2x-6)-\frac{1}{6}ln(x+5)+C$
September 10th 2011, 03:28 AM
HallsofIvy

Re: Partial fractions for integration

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

Quote:

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
$\frac{1}{x^2- 1}= \frac{A}{x- 1}+ \frac{B}{x+1}$

Quote:

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
$\frac{1}{2}ln(\frac{x-1}{x+1})+ C= ln\left(\sqrt{\frac{x-1}{x+1}}\right)+ C$
September 10th 2011, 03:34 AM
celtic1234

Re: Partial fractions for integration

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
September 10th 2011, 03:39 AM
Sudharaka

Re: Partial fractions for integration

Quote:

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.
September 10th 2011, 03:55 AM
celtic1234

Re: Partial fractions for integration

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;

$\frac{5}{4}ln(2x-6)-\frac{5}{8}ln(x+5)$
September 10th 2011, 04:17 AM
Sudharaka

Re: Partial fractions for integration

Quote:

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.

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

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

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

$\frac{5}{8}\ln\left|\frac{x-3}{x+5}\right|+C$
September 10th 2011, 10:22 PM
celtic1234

Re: Partial fractions for integration

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