Question

Неопределенный интеграл.................

Answer 1

$\int x^4 \, \sqrt{1+x^2}\, dx=[\, x=tgt,\; t=arctgx,\; dx=\frac{dt}{cos^2t},\\\\1+x^2=1+tg^2t=\frac{1}{cos^2t}\; ,\; \sqrt{1+x^2}=\frac{1}{cost}\; ,\; x^4=tg^4t\, ]=\\\\=\int \frac{tg^4t\cdot dt}{cost\cdot cos^2t}=\int \frac{cos^4t\cdot dt}{sin^4t\cdot cos^3t}=\int \frac{cost\cdot dt}{sin^4t}=\int (sint)^{-4}\cdot d(sint)=\\\\=[d(sint)=(sint)'dt=cost\, dt\, ]=\frac{(sint)^{-3}}{-3}+C=-\frac{1}{3sin^3t}+C=\\\\=-\frac{1}{3sin^3(arctgx)}+C=-\frac{1}{3\cdot \left (\frac{x}{\sqrt{1+x^2}}\right )^3}+C=-\frac{\sqrt{(1+x^2)^3}}{3x^3}+C$