<s>[INST] Integrate $\int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx $ by differentiating parameter $a$ Can anyone help me calculate the integral using differentiation with respect to a parameter:
$$I(a) = \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx, |a|<1$$
UPD: I try to solve like this:
\begin{align*}
I'(a)&=\int_{0}^{1} \frac{2ax^2}{\sqrt{1-x^2}(1+a^2x^2)} dx
\\&=2a\left(\frac{1}{a^2}\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} - \frac{1}{a^2}\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right)
\\&= 2a\left(\frac{1}{a^2}\arcsin(x)|_0^1-\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right)
\end{align*}
Then I try to solve the second integral:
I am doing a variable change like this:
$$x=\sin(x) ,y=\arcsin(x),dx=\cos(y)dy$$
After that, I solve this integral and finally get:
$$\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}$$
And the problem is when substituting 1 into the result:
$$\lim_{x\to 1}{\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}$$
Nevertheless, this is a difficult way. Is there an easy way?
 [/INST] Note 
$$\lim_{x\to 1}{\frac{\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}= \frac{\arctan^{-1}(\infty)}{\sqrt{1+a^2}}= 
\frac{\frac\pi2}{\sqrt{1+a^2}} $$
Then, you have
$$I’(a)=\frac\pi a\left(1-\frac1{\sqrt{1+a^2}} \right)
$$
and
$$ \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx
= \int_0^a I’(t) dt =\pi \int_0^a \frac1t\left(1-\frac1{\sqrt{1+t^2}} \right) dt \\
= \pi\ln \frac {\sqrt{1+a^2}+1}2
$$
</s>