# Solve. displaystyleintfrac{{sqrt{{x}}}}{{{1}+{sqrt[{{3}}]{{x}}}}}{left.{d}{x}right.}=?

Question
Integrals
Solve. $$\displaystyle\int\frac{{\sqrt{{x}}}}{{{1}+{\sqrt[{{3}}]{{x}}}}}{\left.{d}{x}\right.}=?$$

2020-10-26
Notice that if $$\displaystyle{u}={x}^{{\frac{1}{{6}}}},{t}{h}{e}{n}\ {u}^{2}={x}^{{\frac{1}{{3}}}}{\quad\text{and}\quad}{u}^{3}={x}^{{\frac{1}{{2}}}}$$.
Also we have that $$\displaystyle{d}{u}=\frac{1}{{{6}{x}^{{\frac{5}{{6}}}}}}{\left.{d}{x}\right.}=\frac{1}{{{6}{u}^{5}}}{\left.{d}{x}\right.}$$.
So that,
$$\displaystyle\int\frac{\sqrt{{2}}}{{{1}+{x}^{{\frac{1}{{3}}}}}}{\left.{d}{x}\right.}=\int{6}{u}^{5}\frac{{u}^{3}}{{{1}+{u}^{2}}}{d}{u}={6}\int\frac{{u}^{8}}{{{1}+{u}^{2}}}$$
From here, we can simply use some polynomial division to simplify the quotient.Notice then that:
$$\displaystyle{\frac{{{u}^{8}}}{{{1}+{u}^{2}}}}={u}^{6}-{u}^{4}+{u}^{2}-{1}+{\frac{{{1}}}{{{u}^{2}+{1}}}}$$
Thus, we have the folowing:
$$\displaystyle{6}\int\frac{{u}^{8}}{{{1}+{u}^{2}}}{d}{u}={6}{\left(\int{u}^{6}-{u}^{4}+{u}^{2}-{1}+\frac{1}{{{u}^{2}+{1}}}\right)}{d}{u}$$
So we have the last result:
$$6(\int u^{6}-u^{4}+u^{2}-1+\frac{1}{u^{2}+1})du = \frac{6u^{7}}{7}-\frac{6u^{5}}{5}+\frac{6u^{3}}{3}-6u+6\arctan u)$$
$$\displaystyle\frac{{{6}{x}^{{\frac{7}{{6}}}}}}{{7}}-\frac{{{6}{x}^{{\frac{5}{{6}}}}}}{{5}}+\frac{{{6}{x}^{{\frac{1}{{2}}}}}}{{3}}-{6}{x}^{{\frac{1}{{6}}}}+{6}{ \arctan{{x}}^{{\frac{1}{{3}}}}}$$

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