Solution of $\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$ [closed]
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
Solution 1:
Meh.
$$\int_0^{+\infty} \frac{\ln(1+x^2)}{1 + x^2}\ \text{d}x = \frac{1}{2}\int_{-\infty}^{+\infty} \frac{\ln(1+x^2)}{1 + x^2}\ \text{d}x$$
Now set
$$x=\tan y$$
and you get:
$$-\frac{4}{2} \int^{\pi/2}_0 \ln(\cos y)\ \text{d}y = \pi \ln 2$$
Other References
Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
But I prefer my method.