Задача №1436
Условие
Найти интеграл \(\int\frac{\sqrt{x}dx}{\sqrt{x}-\sqrt[3]{x}}\).
Решение
\[
\int\frac{\sqrt{x}dx}{\sqrt{x}-\sqrt[3]{x}}
=\left[\begin{aligned}& u=\sqrt[6]{x};\,x=u^6.\\& dx=6u^5du.\end{aligned}\right]
=\int\frac{6u^8du}{u^3-u^2}
=6\int\frac{u^6du}{u-1}
=6\int\frac{u^6-1+1}{u-1}du=\\
=6\int\frac{(u-1)\left(u^2+u+1\right)\left(u^3+1\right)+1}{u-1}du
=6\int\left(u^5+u^4+u^3+u^2+u+1+\frac{1}{u-1}\right)du=\\
=u^6+\frac{6u^5}{5}+\frac{3u^4}{2}+2u^3+3u^2+6u+6\ln|u-1|+C=\\
=x+\frac{6\sqrt[6]{x^5}}{5}+\frac{3\sqrt[3]{x^2}}{2}+2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\left|\sqrt[6]{x}-1\right|+C
\]
Ответ:
\(x+\frac{6\sqrt[6]{x^5}}{5}+\frac{3\sqrt[3]{x^2}}{2}+2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln\left|\sqrt[6]{x}-1\right|+C\)