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The alternative to substitution for evaluating \begin{equation} \int \frac{x\arcsin{x}}{\sqrt{1-x^2}}\mathrm{d}x \end{equation} is to integrate-by-parts. Set \begin{align} u=&\arcsin{x},& \mathrm{d}v=&\frac{x}{\sqrt{1-x^2}}\mathrm{dx},\\ \mathrm{d}u=&\frac{1}{\sqrt{1-x^2}}\mathrm{d}x,& v=&-\sqrt{1-x^2}. \end{align} Consequently \begin{equation} \int \frac{x\arcsin{x}}{\sqrt{1-x^2}}\mathrm{d}x=-\sqrt{1-x^2}\arcsin{x}+\int \mathrm{d}x=-\sqrt{1-x^2}\arcsin{x}+x+C \end{equation}
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Thu, 02 Sep 2021 18:54 GMT