# Intergration of fraction

• Jan 16th 2013, 11:17 AM
Saxby
Intergration of fraction
Intergrate -x/(1-x^2)dx, i tried partial fractions but i didn't get a valid answer,

Thanks for any help :)
• Jan 16th 2013, 11:20 AM
MarkFL
Re: Intergration of fraction
I would try a u-substitution:

$\displaystyle u=1-x^2\,\therefore\,du=-2x\,dx$
• Jan 16th 2013, 11:31 AM
Saxby
Re: Intergration of fraction
Yep i've got the answer, -(1/2)ln(1-x^2). Thanks for the suggestion :)
• Jan 16th 2013, 11:46 AM
MarkFL
Re: Intergration of fraction
You don't want the negative sign out front...

$\displaystyle \int\frac{-x}{1-x^2}\,dx=\frac{1}{2}\int\frac{-2x\,dx}{1-x^2}=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u| +C=\frac{1}{2}\ln|1-x^2|+C$
• Jan 16th 2013, 02:22 PM
Soroban
Re: Intergration of fraction
Hello, Saxby!

Quote:

$\displaystyle \int\frac{ -x}{1-x^2}\,dx$

I tried partial fractions but i didn't get a valid answer. .Must have made errors.

We have: .$\displaystyle \int\frac{-x}{1-x^2}\,dx \:=\:\int\frac{x}{x^2-1}\,dx \:=\:\int\frac{x}{(x-1)(x+1)}\,dx$

$\displaystyle \frac{x}{(x-1)(x+1)} \:=\:\frac{A}{x-1} + \frac{B}{x+1} \quad\Rightarrow\quad x \:=\:A(x+1) + B(x-1)$

Let $\displaystyle x = 1\!:\;1 \:=\:A(2) + B(0) \quad\Rightarrow\quad A = \tfrac{1}{2}$

Let $\displaystyle x = \text{-}1\!:\;\text{-}1 \:=\:A(0) + B(\text{-}2) \quad\Rightarrow\quad B = \tfrac{1}{2}$

We have: .$\displaystyle \int\left(\frac{\frac{1}{2}}{x-1} + \frac{\frac{1}{2}}{x+1}\right)\,dx \;=\; \tfrac{1}{2}\int\left(\frac{1}{x-1} + \frac{1}{x+1}\right)\,dx$

. . . . . . . $\displaystyle =\;\tfrac{1}{2}\big(\ln|x-1| + \ln|x+1|\big)+C \;=\;\tfrac{1}{2}\ln|(x-1)(x+1)| + C$

. . . . . . . $\displaystyle =\;\tfrac{1}{2}\ln\left|x^2-1\right| + C$
• Jan 17th 2013, 10:07 AM
zaidalyafey
Re: Intergration of fraction
Hi guys , happy to see you here (Itwasntme)

partial fractions can be made easier ,if we played a little trick

$\displaystyle \frac{1}{2} \int \frac{2x}{x^2-1}\, dx =\frac{1}{2} \int \frac{1+x+x-1}{x^2-1}\, dx=\frac{1}{2} \int \frac{x+1}{x^2-1}+\frac{x-1}{x^2-1}\, dx$

$\displaystyle =\frac{1}{2} \int \frac{1}{x-1}+\frac{1}{x+1}\, dx$
• Jan 17th 2013, 10:43 AM
MarkFL
Re: Intergration of fraction
Hey there! Glad to see a familiar name!

We could avoid the negative sign issue altogether by writing:

$\displaystyle \int\frac{-x}{1-x^2}\,dx=\int\frac{x}{x^2-1}\,dx=\frac{1}{2}\int\frac{2x\,dx}{x^2-1}=\frac{1}{2}\int\frac{du}{u}=$

$\displaystyle \frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|x^2-1|+C$