Задача №2051

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