How to compute $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$ I have first written $x^4-2x^2+2=(x^2-1)^2+1$ then I choose $t=x^2-1$. From this I got $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$
From this how to proceed?
 A: HINT:
$$\frac{\text{d}}{\text{dn}}\left(\int_0^\infty\frac{\ln\left(1+\text{n}x^2\right)}{\sqrt{x}\left(1+x\right)}\space\text{d}x\right)=\int_0^\infty\frac{x^\frac{3}{2}}{\left(1+\text{n}x^2\right)\left(1+x\right)}\space\text{d}x=$$
$$\frac{\pi}{2\left(1+\text{n}\right)}\left(2-\frac{\sqrt{2}\left(\sqrt{\text{n}}-1\right)}{\text{n}^\frac{3}{4}}\right)\tag1$$
A: Let 
\begin{eqnarray}
I(a)=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx
\end{eqnarray}
Then, 
\begin{eqnarray}
I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)}dx\\
&=&\frac{1}{1+a}\int_0^\infty\left(\frac{1}{1+x^2}+\frac{x^2-1}{1+a x^4}\right)dx
=\frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)\\
\end{eqnarray}
Thus,
\begin{align}
&\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt \overset{x=\sqrt t}=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx\\
= & I(1)=\int_0^1 I’(\alpha)d\alpha 
= \int_0^1 \frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)d\alpha \\
= & \frac\pi2(\ln2 + 2\coth^{-1} \sqrt2 )
\end{align}
