積分ボット由来の問題らしいがこんな解き方になってしもうた

$${\int }_0^1\frac{x}{(1-s^2{x}^2)\sqrt{1-x^2}}\log \frac{1-x}{1+x}dx=\frac{\pi }{2s\sqrt{1-s^2}}\log \frac{1-s}{1+s}$$

$$\begin{array}{rl}& {\int }_0^1\frac{x}{(1-s^2{x}^2)\sqrt{1-x^2}}\log \frac{1-x}{1+x}dx\\ & =\frac{1}{2}{\int }_0^1\frac{1}{(1-s^{2}x)\sqrt{1-x}}\log \frac{1-\sqrt{x}}{1+\sqrt{x}}dxルート置換\\ & =\frac{1}{2}{\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}{x}^n\frac{1}{\sqrt{1-x}}\log \frac{1-\sqrt{x}}{1+\sqrt{x}}dx等比数列の和\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}{\int }_0^1\frac{x^n}{\sqrt{1-x}}\log \frac{1-\sqrt{x}}{1+\sqrt{x}}dxべき級数様様の交換\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}{\int }_0^1\frac{(1-x{)}^n}{\sqrt{x}}\log \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}dx\mathrm{King}\mathrm{Property}便利～\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}{\int }_0^1\frac{1}{\sqrt{x}}\log \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-x{)}^{k}dx正義の二項定理\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\int }_0^1{x}^{k-\frac{1}{2}}\log \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}dx有限和は雑に交換できるぅ\\ & =-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k}){\int }_0^1{x}^{k+\frac{1}{2}}\frac{1}{x\sqrt{1-x}}dx部分積分やでぇ\\ & =-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k})B(k+\frac{1}{2},\frac{1}{2})ベータ関数\\ & =-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k})\frac{\mathrm{\Gamma }(k+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(k+1)}ガンマに書き換え～の\\ & =-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k})\frac{\sqrt{\pi }\frac{(2k-1)!!}{2^k}\cdot \sqrt{\pi }}{k!}特殊値適応しーの\\ & =-\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k})\frac{(2k-1)!!}{2^{k}k!}ととのえーの\\ & =-\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}{s}^{2n}\sum _{k=0}^n\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{n}{k})(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}中央二項係数だぁ\\ & =-\frac{\pi }{2}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}\sum _{n=k}^{\mathrm{\infty }}{s}^{2n}(\genfrac{}{}{0ex}{}{n}{k})少々怪しい変形挟んで(数学科の目の前ではできませんね)\\ & =-\frac{\pi }{2(1-s^2)}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}{\left(\frac{s^2}{1-s^2}\right)}^kちょっと特殊なタイプの二項係数の母関数\\ & =-\frac{\pi }{2(1-s^2)}\frac{2\sqrt{1-s^2}}{|s|}{\sinh }^{-1}\frac{|s|}{\sqrt{1-s^2}}※後でここ解説しますね\\ & =-\frac{\pi }{|s|\sqrt{1-s^2}}{\sinh }^{-1}\frac{|s|}{\sqrt{1-s^2}}\\ & =\frac{\pi }{2s\sqrt{1-s^2}}\log \frac{1-s}{1+s}ここまでする必要はないかもしれない\end{array}$$

$$\begin{array}{rl}& \sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}{t}^k\\ & =\frac{1}{\sqrt{t}}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}{t}^{k+\frac{1}{2}}\\ & =\frac{1}{\sqrt{t}}{\int }_0^t\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}{u}^{k-\frac{1}{2}}du\\ & =\frac{1}{\sqrt{t}}{\int }_0^t\frac{1}{\sqrt{u}}\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2k}{k}){(-\frac{u}{4})}^{k}du\\ & =\frac{1}{\sqrt{t}}{\int }_0^t\frac{1}{\sqrt{u}}\frac{1}{\sqrt{1-4\cdot \frac{-u}{4}}}du中央二項係数の母関数\\ & =\frac{2}{\sqrt{t}}{\sinh }^{-1}\sqrt{t}\\ & \therefore \sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k+\frac{1}{2}}(\genfrac{}{}{0ex}{}{2k}{k})4^{-k}{\left(\frac{s^2}{1-s^2}\right)}^k=\frac{2}{\sqrt{\frac{s^2}{1-s^2}}}{\sinh }^{-1}\sqrt{\frac{s^2}{1-s^2}}\\ & =\frac{2\sqrt{1-s^2}}{|s|}{\sinh }^{-1}\frac{|s|}{\sqrt{1-s^2}}\end{array}$$