Question

integral of x/((x^2 +2)^2) from 0 to infinity

Answer 1

1/4

Answer 2

how did you do it?

Answer 3

First, what you need to do is notice that this is an improper integral

then rewrite as

\[ \lim_{a \rightarrow \infty} \int\limits_{0}^{\infty} x/(x^2+2)^2 dx\]

now do u substitution
\[u = x^2+2\]
\[du =2x dx\]
\[du/2 = x dx\]

change the integral with respect to u
\[\lim_{a \rightarrow \infty} 1/2\int\limits_{2}^{a} 1/u^2 du\]

don't forget to alter the integration bounds 0 and a, for a, this doesn't matter as this is representative of infinity. for 0, note that u=x^2 + 2 so the new lower bound is 0^2 +2 = 2

integrate
\[\lim_{a \rightarrow \infty} 1/2\ [-1/a + 1/2] = 1/4\]

note that -1/a where a can be "very large" as it goes to infinity makes -1/a go to zero

I'm new to this site so it took me awhile to get the fancy LaTeX format up. sorry for the delay

Answer 4

oops, in the first equation, note that infinity is supposed to be subbed with a