Question

∫from 0 to ∞ of 1/(x^2+1)(x+1)dx i need an explanation on how to solve

Answer 1

help

Answer 2

\[\int\limits_{0}^{\infty} \frac{ 1 }{ (x^2+1)(x+1) } dx\]

Answer 3

\[\frac{1}{(x^2+1)(x+1)}=\frac 12\left[-\frac{x}{x^2+1}+\frac{1}{x+1}+\frac{1}{x^2+1}\right]\]

use this ...

Answer 4

i've used wolframalpha and it gave me \[\frac{ \Pi }{ 4 }\] as a result but it doesn't tell me the steps. do i split the integral or just resolve it as it is. either way a don't now where that \[\Pi \] is coming from

Answer 5

So you can see that the wolfram alpha gives the answer for the indefinite integral as
\[-\frac 14 \ln(x^2+1)+\frac 12 \ln(x+1)+\frac 12 \tan^{-1}(x)+c\]
to find the integral from \(0\) to \(\infty\),
\[\left[-\frac 14 \ln(x^2+1)+\frac 12 \ln(x+1)+\frac 12 \tan^{-1}(x)\right]_0^\infty \\
=\left[\frac14\{2\ln(x+1)-\ln(x^2+1)\}+\frac 12 \tan^{-1}x\right]_0^\infty\\
=\left[\frac14\ln\left(\frac{(x+1)^2}{x^2+1}\right)+\frac 12 \tan^{-1}x\right]_0^\infty\\
=\left[\frac14\ln\left(\frac{(1+\frac 1x)^2}{1+\frac1 {x^2}}\right)+\frac 12 \tan^{-1}x\right]_0^\infty\\
=\left[\frac 14\ln\left(\frac{1+\frac1\infty }{1+\frac1{\infty^2}}\right)+\frac12\tan^{-1}(\infty)\right]-\left[\frac14\{2\ln(0+1)-\ln(0^2+1)\}+\frac 12 \tan^{-1}0\right]\\
=\left[\frac14\ln(1)+\frac{\pi}{4} \right]-\left[\frac14\{2\ln(1)-\ln(1)\}+\frac 12 (0)\right]\\
=\frac\pi4-0=\frac \pi 4\]