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.

Following @Zacky's hint, use $\frac{t}{1+2t^2}=\frac{2t-t}{1+2t\cdot t}$ to rewrite the integral as the Frullani integral$$\int_0^\infty\frac{\arctan(2t)-\arctan t}{t}dt=(\arctan0-\arctan\infty)\ln\frac12=\frac{\pi}{2}\ln 2.$$
