Answer
$$\int_{0}^{1} \frac{1}{1+\sqrt[3]{x}} d x=-\frac{3}{2}+3 \ln (2) $$
Work Step by Step
Given $$\int_{0}^{1} \frac{1}{1+\sqrt[3]{x}} d x$$ So: let $u^3=x\rightarrow 3u^2du$. With this substitution, the integral can be computed as follows: \begin{aligned} I&=\int_{0}^{1} \frac{1}{1+\sqrt[3]{x}} d x\\ &=\int_{0}^{1} \frac{3 u^{2}}{1+u} \mathrm{d} u \\ &=3\int_{0}^{1} \frac{ u^{2}-1+1}{1+u} \mathrm{d} u \\ &=3 \int_{0}^{1} \frac{u^{2}-1}{1+u}+\frac{1}{1+u} \mathrm{d} u \\ &=3 \int_{0}^{1} \frac{(u -1)(u+1)}{1+u}+\frac{1}{1+u} \mathrm{d} u \\&=3 \int_{0}^{1} u-1+\frac{1}{1+u} \mathrm{d} u \\ &=3\left[\frac{u^{2}}{2}-u+\ln (u+1)\right]_{0}^{1} \\ &=3\left(\frac{1}{2}-1+\ln (2)\right) \\ &=-\frac{3}{2}+3 \ln (2) \end{aligned}
