Задача №1540

\[ \int\frac{x^3dx}{\sqrt{1+2x^2}} =[t=x^2] =\frac{1}{2}\int\frac{tdt}{\sqrt{1+2t}} =\frac{1}{4}\int\frac{(2t+1-1)dt}{\sqrt{2t+1}}=\\ =\frac{1}{8}\int\left((2t+1)^{\frac{1}{2}}-(2t+1)^{-\frac{1}{2}}\right)d(2t+1) =\frac{\sqrt{\left(2x^2+1\right)^3}}{12}-\frac{\sqrt{2x^2+1}}{4}+C =\frac{\left(x^2-1\right)\cdot\sqrt{2x^2+1}}{6}+C \]