Find :
$$I=\int_0^{\infty}\frac{\ln(1+x^b)}{x^2+ax+1}\,dx$$
where $b>0$ , $a\in \Bbb R$.
I don't know if this integral have a closed form but my try as follow : at first I divide integral $\int_0^{1}+\int_1^{\infty}$
I use sub :
$y=\frac{1}{x}$ then $dy=-\frac{1}{x^2}\,dx$.
$$I=\int_0^{1}\mkern-5mu\frac{\ln(1+x^b)}{x^2+ax+1}+\int_0^{1}\!\frac{\ln(1+x^b)-b\ln x}{x^2+ax+1}=2\int_0^{1}\mkern-5mu\frac{\ln(1+x^b)}{x^2+ax+1}-\int_0^{1}\mkern-6mu\frac{\ln x}{x^2+ax+1}$$
But how I complete this work ?