Question

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

Answer 1

let u = x^2+1

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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