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How to prove $\int_{0}^{\infty}\frac{\cos x}{x}\ln{(1+x^{2})} dx = \frac{\pi}{2}$

I would like to show that $$\int_{0}^{\infty}\frac{\cos x}{x}\ln{(1+x^{2})} dx = \frac{\pi}{2}.$$
I got something similar, but I do not know how to prove this. I tried to use integration by parts, but I don’t have a theorem I can use.
Any hint?

A:

hint:
$$\int_{0}^{\infty}\frac{\cos x}{x}\ln{(1+x^{2})} dx=$$
$$\frac 12 \int_{0}^{\infty}\frac{\cos x}{x}\frac{\ln{(1+x^{2})}}{x} dx-\frac 12\int_{0}^{\infty}\frac{\cos x}{x}\frac{\ln{(1+x^{2})}}{x^{2}} dx$$
$$\frac 12 \int_{0}^{\infty}\frac{\cos x}{x}\frac{\ln{(1+x^{2})}}{x} dx-\frac 12\int_{0}^{\infty}\frac{\cos x^{2}}{x^{2}}\frac{\ln{(1+x^{2})}}{x} dx$$

now use the below:
$$\int x^{\alpha}\frac{\ln{(1+x)}}{x} dx=\frac{\alpha}{1+\alpha}x^{\alpha+1}\ln{(1+x)}+\frac{\alpha(\alpha+1)}{(1+\alpha)^{2}}x^{\alpha+1}-\frac{\alpha}{1+\alpha}$$

A:

You can use integration by parts like this
\begin{align}
I=&\frac{1}{2}\int^\

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