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.
 A: 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:
$$I=\int_0^\infty \frac{\arctan(2x)-\arctan x}{x}dx\overset{IBP}=\int_0^\infty \ln x\left(\frac{1}{1+x^2}-\frac{2}{1+4x^2}\right)dx$$
$$2\int_0^\infty \frac{\ln x}{1+4x^2}dx\overset{2x\to x}=\int_0^\infty \frac{\ln x-\ln 2}{1+x^2}dx$$
$$\Rightarrow I=\int_0^\infty \frac{\ln x -\ln x+\ln 2}{1+x^2}dx=\ln 2\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2}\ln 2$$
A: 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$$\int_0^\infty\frac{\arctan(2t)-\arctan t}{t}dt=(\arctan0-\arctan\infty)\ln\frac12=\frac{\pi}{2}\ln 2.$$
A: By integration by parts we have
$$I=\int_0^\infty\ln x\cdot\frac{2x^2-1}{4x^4+5x^2+1}dx$$
$$=\int_0^\infty\frac{\ln x}{1+x^2}dx-\color{red}{\int_0^\infty\frac{2\ln x}{1+4x^2}dx}$$
$$\overset{\color{red}{2x\mapsto x}}{=}\int_0^\infty\frac{\ln2}{1+x^2}dx=\frac{\pi}{2}\ln2$$
A: I found another way to solve this is to make the integral to be a double integral, and change the order. But the key still is $\arctan(\frac{t}{1+2t^2})=\arctan(2t)-\arctan(t)$.
$$\int_0^\infty\frac{\arctan(\frac{t}{1+2t^2})}{t}=\int_0^\infty\frac{\arctan(2t)-\arctan t}{t}dt=\int_0^\infty \frac{1}{t} \int_1^2 \frac{t}{1+(yt)^2} dydt$$
$$=\int_1^2 \int_0^\infty \frac{1}{1+(yt)^2} dtdy=\int_1^2 \frac{1}{y} \arctan(\infty)-\arctan(0) dy=\frac{\pi}{2}ln2$$
