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Applying integration by parts splits the integral into 3 integrals, $\displaystyle \begin{aligned}I&=\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x\\&=-\int_{0}^{1} \sin ^{-1} x \ln (1+x) d\left(\frac{1}{x}\right) \\&=-\left[\frac{\sin ^{-1} x \ln (1+x)}{x}\right]_{0}^{1}+\underbrace{\int_{0}^{1} \frac{\ln (1+x)}{x \sqrt{1-x^{2}}}}_{K} +\underbrace{\int_{0}^{1}\frac{\sin ^{-1} x}{x}}_{L} d x-\underbrace{\int_{0}^{1} \frac{\sin ^{-1} x}{1+x}}_{M} d x \end{aligned} \tag*{} $ Letting $x= \cos \theta$ for $K$ and $\sin^{-1}x \mapsto x$ for $L$ and $M$, yields $\displaystyle I=-\frac{\pi}{2} \ln 2 +\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta}_{K}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{\sin x} d x}_{L}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{1+\sin x} d x }_{M}\tag*{} $


For the integral $ K,$putting $ a=1$ in my post yields $\displaystyle \boxed{K=\frac{\pi^{2}}{8}}\tag*{} $


For the integral $ L,$ integration by parts yields $\displaystyle \begin{aligned}L &=\int_{0}^{\frac{\pi}{2}} x d \ln (\sin x) \\&=[x \ln (\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \\&=\boxed{\frac{\pi}{2} \ln 2}\end{aligned}\tag*{} $


For the integral $ M,$ integration by parts yields $\displaystyle \begin{aligned}M &=\int_{0}^{\frac{\pi}{2}} x d \ln (1+\sin x)\\&=[x \ln (1+\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x) d x \\&=\frac{\pi}{2} \ln 2-\underbrace{\int_0^{\frac{\pi}{2} }\ln (1+\sin x) d x}_{N}\end{aligned}\tag*{} $ For the integral $ N,$ using my post in the second last step yields $\displaystyle \begin{aligned}N \stackrel{x\mapsto\frac{\pi}{2}-x}{=} &\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x \\=&\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\=&\frac{\pi}{2} \ln 2+4\left(\frac{1}{4}(2 G-\pi \ln 2)\right) \\=&\boxed{-\frac{\pi}{2} \ln 2+2 G}\end{aligned}\tag*{} $ where $ G$ is the Catalan’s Constant.


Putting them together yields $\displaystyle \boxed{I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G} \tag*{} $


Question: Is there any shorter solution?

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    $\begingroup$ I am not satisfied with my solution which is lengthy! Just want to improve it. Do you have any suggestion? $\endgroup$
    – Lai
    Commented Apr 15, 2022 at 10:25
  • $\begingroup$ An old issue of RMM, see here ssmrmh.ro/2021/11/02/integral-calculus-567 $\endgroup$
    – Naren
    Commented Apr 16, 2022 at 11:48

2 Answers 2

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A self-contained solution \begin{align} I=&\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x =\int_{0}^{1} \sin ^{-1}x \>d\left( \ln x -\frac{1+x}x \ln (1+x)\right)\\ \overset{ibp} =&\> -\pi \ln 2-{\int_{0}^{1} \frac{\ln \frac x{1+x}-\frac1x \ln(1+x)}{\sqrt{1-x^{2}}}}\> \overset{x=\frac{2t}{1+t^2}}{dx}\\ =&\>-\pi\ln2 + \int_0^1 \underset{=K}{\frac{\ln\frac{(1+t)^2}{1+t^2}}{t}}dt -2\int_0^1 \underset{=J}{\frac{\ln\frac{2t}{(1+t)^2}}{1+t^2}}dt \end{align} with \begin{align} K=&\>\int_0^1 \frac{\ln (1+t)^2}{t}dt -\int_0^1 \frac{\ln (1+t^2)}{t}\overset{t^2\to t}{dt} =\frac32 \int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}8\\ J=& \int_0^1 \frac{\ln t}{1+t^2}dt +\int_0^1 \frac{\ln \frac2{(1+t)^2}}{1+t^2}\overset{t\to\frac{1-t}{1+t}}{dt}= \int_0^1 \frac{\ln t}{1+t^2}dt=-G\\ \end{align} Thus $$I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G$$

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Perhaps not 100% satisfactory, as I've performed each step via Mathematica rather than a step-by-step derivation, but the following could be considered a more simple solution. Note that $$\ln(1+x) = \sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$$ and $$\int_0^1\sin^{-1}x\,x^n dx = \frac{\sqrt{\pi}}{n+1}\left( \frac{\sqrt\pi}{2} - \frac{\Gamma\big(1+\frac n2\big)}{(n+1)\Gamma\big(\frac{n+1}{2}\big)} \right),$$ where, for $n=-1$, the above is understood to be $\frac{\pi\ln2}{2}$. Performing the infinite sum yields the answer given in the original question.

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