calculate $\int_{0}^{\infty}\log^n(x)\log(1+\frac{b^{2}}{x^{2}})dx=? (n\in N^* , b\geq2)$ 
calculate $$\int_{0}^{\infty}(\log(x))^{n}\log(1+\frac{b^{2}}{x^{2}})dx=?  (n\in N^*, b\geq2)$$

for $n=1$:
\begin{align*}
\int_{0}^{\infty}\log(x)\log(1+\frac{b^2}{x^2})dx&=\int_{0}^{\infty}\log(x)\log(\frac{x^2+b^2}{x^2})dx\\
&=\int_{0}^{\infty}\log(x)(\log(x^2+b^2)-\log(x^2))dx\\
&=\int_{0}^{\infty}\log(x)\{\log(x^2+b^2)\}_{0}^{b}dx\\
&=\int_{0}^{\infty}\log(x)\int_{0}^{b}\frac{2t}{x^2+t^2}dtdx\\
&=\int_{0}^{b}\int_{0}^{\infty}\frac{2t\log(x)}{t^2+x^2}dxdt\\
&=\int_{0}^{b}\frac{2t\pi\log(t)}{2t}dt\\
&=\pi\int_{0}^{b}\log(t)dt=\pi\{t\log(t)-t\}_{0}^{b}\\
&=\pi b(\log(b)-1)\\
\end{align*}
Wait for your help $n\geq2$

$$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(1+\frac{b^2}{x^2}\right)\:dx=\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(b^2+x^2\right)\:dx-2\int _0^{\infty }\ln ^{n+1}\left(x\right)\:dx$$
Now consider the following definition of the beta function
$$\int _0^{\infty }\frac{x^{\alpha -1}}{\left(1+x\right)^{\beta }}\:dx=\text{B}\left(\alpha ,\beta -\alpha \right)$$
Now using the sub $x=\frac{x^2}{b^2}$ gets
$$\int _0^{\infty }\frac{x^{2\alpha -1}}{\left(b^2+x^2\right)^{\beta }}\:dx=\frac{1}{2}b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)$$
Using these definitions gets us
$$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(b^2+x^2\right)\:dx=-\frac{1}{2^{n+1}}\lim_{\alpha\rightarrow 1/2 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^n\partial \beta }b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)$$
$$-2\int _0^{\infty }\ln ^{n+1}\left(x\right)\:dx=-2\lim_{\alpha\rightarrow 1 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^{n+1}}\text{B}\left(\alpha ,\beta -\alpha \right)$$
Collecting both results gives us the not so nice generalizaton for your integral
$$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(1+\frac{b^2}{x^2}\right)\:dx$$
$$=-\frac{1}{2^{n+1}}\lim_{\alpha\rightarrow 1/2 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^n\partial \beta }b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)-2\lim_{\alpha\rightarrow 1 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^{n+1}}\text{B}\left(\alpha ,\beta -\alpha \right)$$
Putting this on mathematica for $n=1$ $b=1$
gives me
$$-\frac{1}{4}\left(2\pi \psi \left(-\frac{1}{2}\right)-2\pi \psi \left(\frac{1}{2}\right)\right)-0$$
$$=-\frac{1}{4}\left(4\pi -2\pi \gamma -4\pi \ln \left(2\right)+2\pi \gamma +4\pi \ln \left(2\right)\right)=-\pi $$
Which agrees with the result mentioned by FDP.
