How to evaluate $ \int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $ How can I evaluate
$$ \int_{0}^{1} \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $$
U-substitution has not worked for me. Integration by parts, Differentiation under integral sign, Mathematica is not coming up with a solution either.
Is there a closed form for this integral?
Thank you kindly for your help and time.
 A: To calculate this integral I'll use series expansion $$\frac 1 {\sqrt{1+x^2}}=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n}$$ for $|x|\le1$
$$I=\int_{0}^{1}\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n} \ln\left(x+\sqrt{1-x^2}\right)dx$$
By dominated convergence
$$I=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}
\int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n} dx$$
Let $$ J=\int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n}$$
Now, Let $x=\cos\theta$
$$\implies J=\int_{0}^{\fracπ2}\ln\left(\cos\theta+\sin\theta\right)\left(\cos^{2n}\theta\right) (\sin\theta) d\theta$$
$$ \implies J=\frac12 \int_{0}^{\fracπ2}\ln\left(1+\sin2\theta\right)\left(\cos^{2n}\theta\right)\left(\sin\theta\right) d\theta$$
$$ \implies J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {\sin^k 2\theta}{k} d\theta$$
$$ \implies  J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {2^k \left(\sin^k \theta \right)\left(\cos^k\theta\right)}{k} d\theta$$
By dominated convergence
$$J=  \sum_{k=1}^{\infty }\frac {(-1)^{k-1} 2^{k-1}}{k}\int_{0}^{\fracπ2}\left(\cos^{2n+k}\theta\right) \left(\sin^{k+1}\theta\right) d\theta$$
Using $$\int_{0}^{\fracπ2}\left(\sin^m\theta\right) \left(\cos^n\theta\right)d\theta=\frac{\Gamma\left(\frac{n+1}2\right)
\Gamma\left(\frac{m+1}2\right)}{2
\Gamma\left(\frac{m+n+2}2\right)}$$
$$J=\sum_{k=1}^{\infty }\frac{(-1)^{k-1} 2^{k-2}}{k}\frac{\Gamma\left(\frac{k+2}2\right)
\Gamma\left(\frac{2n+k+1}2\right)}{
\Gamma\left(\frac{2n+2k+3}2\right)}$$
On substituting $J$ in orignal integral, we get
$$I=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac {(-1)^{(n+k-1)}}{2^{(2n-k+2)}k}\binom{2n}{n}\frac{\Gamma\left(\frac{k+2}2\right)
\Gamma\left(\frac{2n+k+1}2\right)}{
\Gamma\left(\frac{2n+2k+3}2\right)}$$
