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?
 A: A slightly more elementary approach to the same binomial series is possible.  Consider $$I = \int_{x=0}^1 \frac{\log (1+x^2)}{\sqrt{1-x^2}} \, dx.$$  The substitution $$x = \sin \theta, \quad dx = \cos \theta \, d\theta$$ gives $$I = \int_{\theta=0}^{\pi/2} \log (1 + \sin^2 \theta) \, d\theta = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \int_{\theta=0}^{\pi/2} \sin^{2k} \theta \, d\theta.$$  In another answer we proved using only integration by parts the identity $$\int_{\theta=0}^{\pi} \sin^{2k} \theta \, d\theta = \binom{2k}{k} \frac{\pi}{4^k},$$  thus $$I = -\frac{\pi}{2} \sum_{k=1}^\infty \binom{2k}{k} \frac{\left(-1/4\right)^k}{k},$$ and the rest of the computation follows.
A: In the OP's comments he asks if integrating the function on $[0,1]$ yields a simple closed form. In this answer I prove that the integral on this interval equals $\pi \log\left(\frac{1}{2} + \frac{1}{\sqrt{2}}\right)$
$$\int_0^1 \frac{\log(1+x^2)}{\sqrt{1-x^2}}dx = \int_0^1 \left(\frac{1}{\sqrt{1-x^2}} \sum_{n=1}^\infty (-1)^{n+1}\frac{x^{2n}}{n}\right)dx$$
We now move some things around and interchange the integral and summation to get
$$\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}dx\right) \tag{1}$$
We recognize the inner integral as a Beta Function. We show this by letting $x = \sin(u)$
$$\int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}dx = \int_0^1 \frac{\sin^{2n}(u)}{\sqrt{1-\sin^2u}}\cos(u)du = \int_0^{\pi/2} \sin^{2n}(u)\,du = \frac{1}{2}B(n+1/2,1/2)$$
Rewriting this in terms of the Gamma function and combining terms, we can replace $(1)$ with
$$\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}}{n} \frac{\sqrt{\pi} \,\Gamma(n + 1/2)}{2\,\Gamma(n + 1)}\right) = \frac{\pi}{2}\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}}{n} \left(\frac{1}{4}\right)^n{2n \choose n}\right) \tag{2}$$  

At this point we look at a different summation for a minute: the binomial series
$$\sum_{n=0}^\infty {2n \choose n} t^n = \frac{1}{\sqrt{1-4t}} \implies \sum_{n=1}^\infty {2n \choose n} t^{n-1} = \frac{1-\sqrt{1-4t}}{\sqrt{1-4t}}$$
We can now integrate to find that
$$\int_0^x\sum_{n=1}^\infty {2n \choose n} t^{n-1}dt = \int_0^x \frac{1-\sqrt{1-4t}}{t\sqrt{1-4t}} dt$$
Interchanging the integral and sum gives us (after relatively easy integrations)
$$\sum_{n=1}^\infty \frac{{2n \choose n}}{n} x^n = \log(4)-2\log(1+\sqrt{1-4x})$$
Letting $x = \frac{-1}{4}$ and multiplying both sides by $-1$, we get
$$-\sum_{n=1}^\infty \frac{{2n \choose n}}{n} \left(\frac{-1}{4}\right)^n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \left(\frac{1}{4}\right)^n {2n \choose n}= 2\log(1+\sqrt{2})-2\log(2)$$

We have now evaluated $(3)$; accordingly, we can now prove the initial claim.
$$\frac{\pi}{2}\sum_{n=1}^\infty \left(\frac{(-1)^{n+1}}{n} \left(\frac{1}{4}\right)^n{2n \choose n}\right) = \pi(\log(1+\sqrt{2})-\log(2)) = \pi \log\left(\frac{1}{2} + \frac{1}{\sqrt{2}}\right) \quad \square$$
A: 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}
