Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$

Another method which gives a closed form result. Using the identity \begin{equation} \ln\frac{1-x}{1+x}=-2\operatorname{arctanh} x \end{equation} and parity of the integrand, the integral can be written as \begin{align} I(k) &= \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\\ &=-\int_{-1}^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\operatorname{arctanh} x\\ &=-\int_{-\infty}^\infty \frac{u\sinh u}{\cosh^2u-k^2\sinh^2u}du \end{align} The last integral is obtained by the substitution $x=\tanh u$. Then \begin{align} I(k)&=-\frac{1}{2k}\left[J(k)-J(-k)\right]\\ J(k)&=\int_{-\infty}^\infty \frac{udu}{\cosh u-k\sinh u} \end{align}

To evaluate $J(k)$, we enforce the substitution $t=e^u$ and, denoting $m^2=\frac{1+k}{1-k}$, we obtain \begin{align} J(k)&=\frac{2}{1-k}\int_0^\infty \frac{\ln v}{v^2+m^2}dv\\ &=\frac{2}{m(1-k)}\int_0^\infty \frac{\ln mt}{t^2+1}dt\\ &=\frac{2}{m(1-k)}\left[\int_0^\infty \frac{\ln t}{t^2+1}dt+\int_0^\infty \frac{\ln m}{t^2+1}dt\right] \end{align} We know that $\int_0^\infty \frac{\ln t}{t^2+1}dt=0$, thus \begin{equation} J(k)=\frac{\pi \ln m}{m(1-k)}=\frac{\pi}{2}\frac{\ln \frac{1+k}{1-k}}{\sqrt{1-k^2}} \end{equation} Finally, \begin{equation} I(k)=\frac{\pi}{2k\sqrt{1-k^2}}\ln\frac{1- k}{1+ k} \end{equation}

A great approach on your part! Let me try another.

$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} \int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2} $$

So we really only need to solve the integral in the form:

$$Y(k,n+1)=\int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$

Note that:

$$Y(k,0)=\int_0^1 \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)}=\frac{\pi}{2 \sqrt{1-k^2}}$$

This can be seen directly from the antiderivative (check by differentiation):

$$ \int \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)} =\frac{1}{\sqrt{1-k^2}} \tan^{-1} \left( \frac{\sqrt{1-k^2} x}{\sqrt{1-x^2}} \right)$$

Let's try something for the general case:

$$Y(k,n+1)=\frac{1}{k^2}\int_0^1 \frac{x^{2n}(k^2x^2-1+1)}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}=$$

$$=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$

From which we write a recurrence relation:

$$Y(k,n+1)=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} Y(k,n)$$

Let us denote:

$$P(n)=\int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}$$

Then:

$$Y(k,n+1)=-\frac{1}{k^2} \left( P(n)-Y(k,n) \right)$$

$P(n)$ is just Beta function and has the form:

$$P(n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!}$$

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In other words, while this solution is by no means complete, we already obtained:

The series representation of the integral:

$$I(k)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} Y(k,n+1) \tag{1}$$

The recurrence relation together with the initial condition, which allows us to compute all the terms:

$$Y(k,n+1)=\frac{1}{k^2} \left(Y(k,n)- \frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} \right) \tag{2}$$

$$Y(k,0)=\frac{\pi}{2 \sqrt{1-k^2}} \tag{3}$$

I will check how this works numerically in a couple of days.

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As can be seen from the integral, $Y(k,n)$ actually has a solution in terms of Gauss Hypergeometric function:

$$Y(k,n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} {_2 F_1} \left(1, n+\frac{1}{2};n+1; k^2 \right)$$

But the recurrence relation is a more simple way to obtain the values without special software. $P(n)$ can be converted into the form of rational multiples of $\pi$.

Final remark: while this doesn't seem to lead us to a closed form, let me note that this approach works for a large class of integrals, as long as we can expand the non-algebraic function as a series.

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Trying to derive a rational form for $P(n)$:

$$ \Gamma \left(n+\frac{1}{2} \right)=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{1}{2} \Gamma \left(\frac{1}{2} \right)=$$

$$=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{\sqrt{\pi}}{2}= $$

$$ =\left(2n-1 \right)\left(2n-3 \right) \cdots 3 \cdot \frac{\sqrt{\pi}}{2^n} $$

So:

$$P(n)=\frac{\pi}{2} \frac{1 \cdot 3 \cdots (2n-3)(2n-1) }{2^n~n!}$$

Separate case: $P(0)= \pi/2$.

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Edit 2

I have checked the algorithm numerically, and it works, but the convergence is terrible, and a huge loss of significance is involved.

Nice approach so far. About the green integrals, you may simply expand $\frac{1}{x\pm\sigma}$ as a geometric series and exploit $\int_{0}^{1} x^{m-1}\log(x)\,dx =-\frac{1}{m^2} $ to get $$ \int_{0}^{1}\frac{-\log x}{\sigma-x}\,dx =\frac{1}{\sigma}\sum_{m\geq 1}\frac{1}{\sigma^m m^2}=\frac{1}{\sigma}\,\text{Li}_2\left(\frac{1}{\sigma}\right)$$ and similarly $$ \int_{0}^{1}\frac{-\log x}{\sigma+x}\,dx =\frac{1}{\sigma}\,\text{Li}_2\left(-\frac{1}{\sigma}\right).$$ Then the reflection formulas for the dilogarithm might induce some extra simplification.
They did it here, for instance.