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Question Number 64382 by Chi Mes Try last updated on 17/Jul/19

$$pleasehelpwithworkings$$$$$$$$\int Ln[\sqrt{}(1-x)+\sqrt{}(1+x)]dx$$

Answered by MJS last updated on 17/Jul/19

$$\mathrm{first}\mathrm{step}:\mathrm{by}\mathrm{parts}$$$$u^{\prime }=1\to u=x$$$$v=\ln (\sqrt{1-x}+\sqrt{1+x})\to v^{\prime }=\frac{1}{2x}-\frac{1}{2x\sqrt{1-x^2}}$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }=x\ln (\sqrt{1-x}+\sqrt{1+x})-\int (\frac{1}{2}-\frac{1}{2\sqrt{1-x^2}})dx$$$$\mathrm{second}\mathrm{step}:\mathrm{standard}\mathrm{integrals}$$$$-\frac{1}{2}\int dx+\frac{1}{2}\int \frac{dx}{\sqrt{1-x^2}}=-\frac{x}{2}+\frac{1}{2}\arcsin x$$$$$$$$\int \ln (\sqrt{1-x}+\sqrt{1+x})dx=-\frac{x}{2}+x\ln (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}\arcsin x+C$$

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