# Math Help - Intergration of fraction

1. ## 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

2. ## Re: Intergration of fraction

I would try a u-substitution:

$u=1-x^2\,\therefore\,du=-2x\,dx$

3. ## Re: Intergration of fraction

Yep i've got the answer, -(1/2)ln(1-x^2). Thanks for the suggestion

4. ## Re: Intergration of fraction

You don't want the negative sign out front...

$\int\frac{-x}{1-x^2}\,dx=\frac{1}{2}\int\frac{-2x\,dx}{1-x^2}=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u| +C=\frac{1}{2}\ln|1-x^2|+C$

5. ## Re: Intergration of fraction

Hello, Saxby!

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

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

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

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

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

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

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

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

. . . . . . . $=\;\tfrac{1}{2}\ln\left|x^2-1\right| + C$

6. ## Re: Intergration of fraction

Hi guys , happy to see you here

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

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

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

7. ## Re: Intergration of fraction

Hey there! Glad to see a familiar name!

We could avoid the negative sign issue altogether by writing:

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

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