Evaluation of $\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$ 
Evaluation of $$\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$$

$\bf{My\; Try:::}$ Let $$I(a) = \int_{0}^{\sqrt{2}-1}\frac{\ln(1+ax^2)}{1+x}dx$$
Now $$I'(a) = \int_{0}^{\sqrt{2}-1}\frac{x^2}{(1+ax^2)(1+x)}dx = \frac{1}{a}\int_{0}^{\sqrt{2}-1}\frac{(1+ax^2)-1}{(1+ax^2)(1+x)}dx$$
So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2}\int_{0}^{\sqrt{2}-1}\frac{1}{(x^2+\frac{1}{a})(1+x)}dx$$
So $$I'(a) = \frac{\ln 2}{2a}-\frac{1}{a^2(a+1)}\int_{0}^{\sqrt{2}-1}\left[\frac{1-x}{x^2+\frac{1}{a}}-\frac{1}{1+x}\right]dx$$
Now How can i solve after that, Help required, Thanks

Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find:
$$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx
\\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx 
\\&=\frac12 I + \frac12 \int_{\sqrt{2}-1}^1 \frac{\ln(1+x^2)}{1+x} dx +\frac12 \int_{\sqrt{2}-1}^1 \frac{\ln2 - 2 \ln(1+x)}{1+x} dx
\\&=\frac12 \int_0^1 \frac{\ln(1+x^2)}{1+x} dx -\frac18 \ln^2 2
\\&= \frac{\ln^2 2}{4}-\frac{\pi^2}{96}.\end{align}$$
You can check the evaluation $$\int_0^1 \frac{\ln(1+x^2)}{1+x} dx =\frac34 \ln^2 2- \frac{\pi^2}{48}$$ here, for example.
