Evaluate $\int \frac{\log(1+x^2)}{\sqrt{1-x^2}} \,\mathrm dx$ 
$\def\d{\mathrm{d}}$Evaluate $$\int \frac{\log(1+x^2)}{\sqrt{1-x^2}}\,\d x.$$

I have used Integration by parts as follows:
$$I=\log(1+x^2) \: \sin^{-1}x-\int \frac{2x \sin^{-1}x}{1+x^2}\,\d x=\log(1+x^2) \: \sin^{-1}x-J,$$ where
$$J=\int \frac{2x \sin^{-1}x}{1+x^2}\,\d x$$ In this put $x=\sin y$ we get
$$J=\int \frac{y \sin2y}{1+\sin^2y}\,\d y.$$
Any clue here?

We can also do this.
Consider the double integral:
\begin{align}I=\int_{0}^{1}\int_{0}^{1} \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)} \ dy \ dx. \end{align}
Integrating with respect to $y$ first, we have 
\begin{align}I=\int_{0}^{1} \frac{\ln(1+x^2)}{\sqrt{1-x^2}} \ dx. \end{align}
On the other hand, we reverse the order of integration 
\begin{align}I=\int_{0}^{1}\int_{0}^{1} \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)} \ dx \ dy. \end{align}
For this, we use the antiderivative of $ \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)}$ is \begin{align} \frac{2}{y} \left(\sin^{-1}(x) - \frac{\tan^{-1} \left( \frac{x\sqrt{1+y^2}}{\sqrt{1-x^2}}\right)}{\sqrt{y^2+1}}  \right), \end{align} 
so we upon plugging in endpoints of integration that 
\begin{align} I= \int_{0}^{1} \frac{\pi}{y} - \frac{\pi}{y\sqrt{1+y^2}} \ dy = \int_{0}^{1} \frac{\pi \sqrt{y^2+1} - \pi}{y\sqrt{1+y^2}} \ dy= \pi \int_{0}^{1}\frac{ y}{\sqrt{1+y^2}(1+\sqrt{1+y^2})} \ dy, \end{align} which we get by simplifying with common denominators and multiplying by the conjugate of the numerator.  Another u-substitution $u=1+\sqrt{1+y^2}, \ du = \frac{y}{\sqrt{1+y^2}(1+\sqrt{1+y^2})}$ gives us
\begin{align} I= \pi \ln(1+\sqrt{2})-\pi\ln(2)  \end{align}
