Задача №1397

\[ \int\frac{\arcsin{x}}{\sqrt{x+1}}dx =\left[\begin{aligned}& u=\arcsin{x};\,du=\frac{dx}{\sqrt{1-x^2}}.\\& dv=\frac{dx}{\sqrt{x+1}}=(x+1)^{-\frac{1}{2}}d(x+1);\,v=2\sqrt{1+x}.\end{aligned}\right]=\\ =2\sqrt{x+1}\arcsin{x}-2\int\frac{\sqrt{1+x}}{\sqrt{1-x^2}}dx =2\sqrt{x+1}\arcsin{x}-2\int\frac{\sqrt{1+x}}{\sqrt{(1-x)(1+x)}}dx=\\ =2\sqrt{x+1}\arcsin{x}-2\int\frac{\sqrt{1+x}}{\sqrt{1-x^2}}dx =2\sqrt{x+1}\arcsin{x}-2\int\frac{dx}{\sqrt{1-x}}=\\ =2\sqrt{x+1}\arcsin{x}+2\int(1-x)^{-\frac{1}{2}}d(1-x) =2\sqrt{x+1}\arcsin{x}+4\sqrt{1-x}+C \]