Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$
My Attempt
Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$
$$
\begin{align}
&\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\
&=\int\limits_0^{\pi/4}\tan y.y^2dy+\int\limits_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-\int2y.\log|\sec y|dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-y^2.\log|\sec y|+\int\frac{\tan y\sec y}{\sec y}y^2dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
\end{align}
$$
How do I proceed further and solve the integration?
Best Answer
Let's try integration by parts: \begin{align} \int x(\arctan x)^2\,dx &= \frac{x^2}{2}(\arctan x)^2- \int\frac{x^2}{2}2\arctan x\frac{1}{1+x^2}\,dx \\ &= \frac{x^2}{2}(\arctan x)^2- \int\frac{1+x^2-1}{1+x^2}\arctan x\,dx \\ &= \frac{x^2}{2}(\arctan x)^2- \int\arctan x\,dx+ \int\frac{1}{1+x^2}\arctan x\,dx \end{align} Now note that $$ \int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx $$ and that $$ \int\frac{1}{1+x^2}\arctan x\,dx= \frac{1}{2}(\arctan x)^2 $$