## a solution to another integral

Posted: October 8, 2014 in Mathematics

This week’s integral:

$\displaystyle \int_0^1 \frac{\log(x+1)}{x^2+1} \;dx$

Solution:

At first glance the integral looks scary because there is no way of making the numerator being the derivative of anything in the denominator. So what can we do about it? How about trying to simplify the denominator or get rid of it completely. Let’s make use of the trig sub,

$x=\tan\theta, \quad dx=\sec^2{\theta}\;d\theta, \quad \tan^2\theta+1=\sec^2\theta$

then we have,

$\displaystyle \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta$

Now apply the trick of

$\displaystyle \int_0^a f(t) \;dt = \int_0^a f(a-t) \;dt$

we have,

$I=\displaystyle \int_0^{\pi/4}\log\left(\tan\theta+1\right)\;d\theta = \int_0^{\pi/4}\log\left(\tan\left(\frac{\pi}{4}-\theta\right)+1\right)\;d\theta$

$\displaystyle I = \int_0^{\pi/4}\log\left(\frac{1-\tan\theta}{1+\tan\theta}+1\right)\;d\theta = \int_0^{\pi/4}\log\frac{2}{1+\tan\theta}\;d\theta$

$\displaystyle I = \int_0^{\pi/4} \log 2 \;d\theta - \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta$

$\displaystyle \therefore\;I = \frac{\log 2}{2}\cdot\frac{\pi}{4} = \frac{\pi\log2}{8}$