Question

Интегралы
Пожалуйста напишите помимо решения примененную формулу или способ

Answer 1

Ответ:

$\frac{6(1+\sqrt[3]{x})^{\frac{7}{2}}}{7}-\frac{18(1+\sqrt[3]{x})^{\frac{5}{2}}}{5}+6(1+\sqrt[3]{x})^{\frac{3}{2}}-6\sqrt{(1+\sqrt[3]{x})}+3C$

Пошаговое объяснение:

$\int {\frac{\sqrt[3]{x}}{\sqrt{1+\sqrt[3]{x}}}}dx\\\\t=\sqrt[3]{x}\\\\dt=\frac{dx}{3\sqrt[3]{x^2}}\\\\\sqrt[3]{x^2}=t^2\\\\dx=3t^2dt\\\\\int {\frac{\sqrt[3]{x}}{\sqrt{1+\sqrt[3]{x}}}}dx=\int {\frac{3t^3dt}{\sqrt{1+t}}}\\\\u=1+t\\\\du=dt$

$\int {\frac{3t^3dt}{\sqrt{1+t}}}=3\int{\frac{(u-1)^3du}{\sqrt{u}}}=3\int{\frac{u^3-3u^2+3u-1}{u^{\frac{1}{2}}}du}=\\\\3\left[\int{\frac{u^3du}{u^{\frac{1}{2}}}}}-\int{\frac{3u^2du}{u^{\frac{1}{2}}}}}+\int{\frac{3udu}{u^{\frac{1}{2}}}}}-\int{\frac{du}{u^{\frac{1}{2}}}}}\right]=3\left[\int{u^{\frac{5}{2}}du}-\int{3u^{\frac{3}{2}}du}+\int{3u^{\frac{1}{2}}du}-\int{u^{-\frac{1}{2}}du}\right]=3\left[\frac{2u^{\frac{7}{2}}}{7}-\frac{6u^{\frac{5}{2}}}{5}+2u^{\frac{3}{2}}-2\sqrt{u}+C\right]\\\\u=1+t=1+\sqrt[3]{x}$$3\left[\frac{2u^{\frac{7}{2}}}{7}-\frac{6u^{\frac{5}{2}}}{5}+2u^{\frac{3}{2}}-2\sqrt{u}+C\right]=\frac{6(1+\sqrt[3]{x})^{\frac{7}{2}}}{7}-\frac{18(1+\sqrt[3]{x})^{\frac{5}{2}}}{5}+6(1+\sqrt[3]{x})^{\frac{3}{2}}-6\sqrt{(1+\sqrt[3]{x})}+3C$