$$\int^\infty_0\frac{\log\left(x^2+1

\right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi

\log^2(2)$$

**Lemma**

$$\int^\infty_0 \frac{\log^3(1 + x^2)}{x^2}\,dx = \pi^3+ 3 \pi \log^2(4)$$

Start by the following

$$\int^{\infty}_0 x^{-p}(1+x)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{\Gamma(1-s)}$$

Let \( x\to x^2 \)

$$\int^{\infty}_0 x^{-2p+1}(1+x^2)^{s-1} dx= \frac{\Gamma(1-p)\Gamma(p-s)}{2\Gamma(1-s)}$$

Let \( p = 3/2 \)

$$\int^{\infty}_0 \frac{1}{x^2(1+x^2)^{1-s}} dx= \frac{\Gamma(-1/2)\Gamma(3/2-s)}{2\Gamma(1-s)}$$

By differentiation

$$\int^\infty_0 \frac{\log^3(1+x^2)}{x^2}\,dx = \frac{\Gamma(-1/2)}{2} \left [ \frac{d^3}{ds^3}\frac{\Gamma(3/2-s)}{2\Gamma(1-s)} \right ]_{s = 1} = \pi^3+ 3 \pi \log^2(4) $$

Now, Consider the function

$$f(z) = \frac{\log^3(1-iz)}{z^2}$$

Define the principle logarithm as follows

$$\log z = \log|z| + i\mathrm{Arg}(z) \,\,\, $$

Note that for \( x > 0 \) the argument can be evaluated as

$$\log z = \log\sqrt{x^2+y^2} + i\arctan(y/x) $$

For the principle logarithm the branch cut is defined as \(\Im(1-iz) =0 ,\Re(1-iz) \leq 0 \) which then reduces for \( z = x+iy \) to \(x=0 , 1+ y \leq 0\) . Hence the branch cut on the imaginary axis where \( x=0 , y < -1 \) . Also note that \( f(z) \) is analytic on the punctured plane since around \( z =0 \)

$$\frac{\log^3(1-iz)}{z^2} = -\frac{((iz)+(iz)^2/2+(iz)^3/3 +\mathcal{O}(z^4))^3}{z^2} =\mathcal{O}(z^2) $$

Hence at \( z = 0 \) we have a removable singularity. Define the following contour were we avoid the branch cut

$$\int_{C_R}f(z)\,dz+\int_{-R}^{0}\frac{\left(\log(\sqrt{1+x^2})+i\arctan(x)\right)^3}{x^2}dx + \int_{0}^R\frac{(\log(\sqrt{1+x^2})+i\arctan(x))^3}{x^2}dx= 0$$

$$\int_{C_R}f(z)\,dz+\int_{0}^{R}\frac{\left(\log(\sqrt{1+x^2})-i\arctan(x)\right)^3}{x^2}dx + \int_{0}^R\frac{(\log(\sqrt{1+x^2})+i\arctan(x))^3}{x^2}dx= 0$$

Note that

$$(x+y)^3 = x^3+y^3+3x^2y+3yx^2$$

This simplifies to

$$\int_{C_R}f(z)\,dz+\int_{0}^{R}\frac{-3 \arctan^2(x) \log(1 + x^2) + 1/4 \log^3(1 + x^2)}{x^2}\,dx= 0$$

Taking the limit

$$\left|\int_{C_R}f(z)\,dz \right| = \left|\int_{C_R}\frac{\log^3(1-iz)}{z^2}\,dz \right| \leq \pi R \frac{(\log |1+R| + 2\pi )^3}{R^2} \sim_{\infty} 0$$

So this simplifies to

$$\int_{0}^{\infty}\frac{ \arctan^2(x) \log(1 + x^2)}{x^2} \,dx = \frac{1}{12} \int^\infty_0 \frac{\log^3(1 + x^2)}{x^2}\,dx$$

Using the Lemma we reach our result.

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