\[∫1/√(1+∛x) dx\]
By Substitution
\begin{array}{c}
\text { Suppose } \begin{aligned}
\Rightarrow z &=\sqrt[3]{x} \\
z^{3} &=x
\end{aligned} \\
3 z^{2} d z=d x \\
\int \frac{3 z^{2}}{\sqrt{1+z}} \mathrm{~d} z
\end{array}
By Substitution
\[Suppose⇒u=√(1+z)\]
\begin{array}{l} u^{2}=1+z \quad, \quad z^{2}=\left(u^{2}-1\right)^{2} \\ 2 u d u=d z \\ \int \frac{3\left(u^{2}-1\right)^{2}}{u} d u \\ 3 \int \frac{u^{4}-2 u^{2}+1}{u} d u \\ 3 \int u^{3}-2 u+\frac{1}{u} d u \\ 3\left(\frac{u^{4}}{4}-u^{2}+\ln u\right)+c \\ 3\left(\frac{(\sqrt{1+z})^{4}}{4}-(\sqrt{1+z})^{2}+\ln |\sqrt{1+z}|\right)+c \\ 3\left(\frac{(\sqrt{1+\sqrt[3]{x}})^{4}}{4}-(\sqrt{1+\sqrt[3]{x}})^{2}+\ln |\sqrt{1+\sqrt[3]{x}}|\right)+c \end{array}
Tags:
integration