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Let the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact
that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain
\begin{align*}
6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \s...
From Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have
$$
\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}
$$
Proof
Integrating $(1+z)^p z^q...7
For the first one,
\begin{align}
\int^1_0\frac{\arctan^2{x}}{x^2}{\rm d}x
=&\color{#BF00FF}{\int^\frac{\pi}{4}_0x^2\csc^2{x}\ {\rm d}x}\\
=&-x^2\cot{x}\Bigg{|}^\frac{\pi}{4}_0+2\int^\frac{\pi}{4}_0x\cot{x}\ {\rm d}x\\
=&-\frac{\pi^2}{16}+4\sum^\infty_{n=1}\int^\frac{\pi}{4}_0x\sin(2nx)\ {\rm d}x\\
=
Evaluate
$$\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx$$
