improper integral: int [0, infty] [ x / (x^2 + 2)^2 ] dx

[summergrl]

Integral from 0 to infinity:
x/(x^2+2)^2

Okay so i know i have to turn it into a limit and then I made u=x^2+2. So it became the limit as b approaches infinity 1/2 the integral from 0 to b of du/u^2. Is that right because then when i try to solve i dont know if it works out right.

 

[galactus]

\(\displaystyle \L\\\int_{0}^{\infty}\frac{x}{(x^{2}+2)^{2}}dx=\frac{-1}{2x^{2}+4}\)

\(\displaystyle \L\\\lim_{x\to\infty}\frac{-1}{2x^{2}+4}=0\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{-1}{2x^{2}+4}=\frac{-1}{4}\)

0-(-1/4)=1/4

 

[summergrl]

\(\displaystyle \L\\\int_{0}^{\infty}\frac{x}{(x^{2}+2)^{2}}dx=\frac{-1}{2x^{2}+4}\)

\(\displaystyle \L\\\lim_{x\to\infty}\frac{-1}{2x^{2}+4}=0\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{-1}{2x^{2}+4}=\frac{-1}{4}\)

0-(-1/4)=1/4

how do you get 1/2x^2+4?

 

[galactus]

u-subbing:

Let \(\displaystyle u=x^{2}+2, \;\ du=2xdx, \;\ \frac{du}{2}=xdx\)

\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{u}du=\frac{-1}{2u}\)

Resub and get:

\(\displaystyle \L\\\frac{-1}{2(x^{2}+2)}=\frac{-1}{2x^{2}+4}\)