How to evaluate $\int_{0}^{\infty}\frac{1}{t}\arctan\left(\frac{t}{1+2t^2}\right)\,\mathrm dt$? I entered this integral into Wolframalpha, and got $$\int_{0}^{\infty}\frac{1}{t}\arctan\left(\frac{t}{1+2t^2}\right)\,\mathrm dt=\frac{1}{2}\pi\log{2}.$$ But it doesn't provide step by step solution for this integral.
This integral is a bonus challenge in my Calculus class, and the professor that the key is $\arctan$. But I don't know is there any special about $$\arctan\left(\frac{t}{1+2t^2}\right),$$ so I tried some common integration method, and it doesn't work.

First notice that:
$$\arctan\left(\frac{x}{1+2x^2}\right)=\arctan\left(\frac{2x-x}{1+2x\cdot x}\right)=\arctan(2x)-\arctan(x)$$
So the integral can be rewritten as:
$$I=\int_0^\infty \frac{\arctan(2x)-\arctan x}{x}dx\overset{IBP}=\int_0^\infty \ln x\left(\frac{1}{1+x^2}-\frac{2}{1+4x^2}\right)dx$$
$$2\int_0^\infty \frac{\ln x}{1+4x^2}dx\overset{2x\to x}=\int_0^\infty \frac{\ln x-\ln 2}{1+x^2}dx$$
$$\Rightarrow I=\int_0^\infty \frac{\ln x -\ln x+\ln 2}{1+x^2}dx=\ln 2\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2}\ln 2$$
