evaluate the integral from negative infinity to infinity of 2x/(x^2+1)^2 dx

Question

zzr0ck3r · Answer

let u = x^2+1

anonymous · Answer

$$I=\int\limits_{\infty}^{\infty}\left( x^2+1 \right)^{-2}*2x~dx$$

zzr0ck3r · Answer

$u=x^2+1\implies du=2x dx\implies dx=\frac{du}{2x}$now plug that in
$\int_{-\infty}^\infty2x\frac{1}{u^2}\frac{du}{2x}=\int_{-\infty}^\infty\frac{1}{u^2}du\=\frac{-2}{u^3}|_{-\infty}^\infty=\frac{-2}{\infty}-(\frac{-2}{-\infty})=0$

zzr0ck3r · Answer

some abuse of notation, but you get the point....

anonymous · Answer

$$\int\limits \frac{ 1 }{ u^2 }du=-\frac{ 1 }{u }$$

zzr0ck3r · Answer

oh woops, i took the derivative.....same answer:)

zzr0ck3r · Answer

$\int_{-\infty}^\infty2x\frac{1}{u^2}\frac{du}{2x}=\int_{-\infty}^\infty\frac{1}{u^2}du\=\frac{-1}{u}|_{-\infty}^\infty=\frac{-1}{\infty}-(\frac{-1}{-\infty})=0$