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

Solve. $\int \frac{\sqrt{x}}{1+\sqrt[3]{x}}dx=?$
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casincal
Notice that if .
Also we have that $du=\frac{1}{6{x}^{\frac{5}{6}}}dx=\frac{1}{6{u}^{5}}dx$.
So that,
$\int \frac{\sqrt{2}}{1+{x}^{\frac{1}{3}}}dx=\int 6{u}^{5}\frac{{u}^{3}}{1+{u}^{2}}du=6\int \frac{{u}^{8}}{1+{u}^{2}}$
From here, we can simply use some polynomial division to simplify the quotient.Notice then that:
$\frac{{u}^{8}}{1+{u}^{2}}={u}^{6}-{u}^{4}+{u}^{2}-1+\frac{1}{{u}^{2}+1}$
Thus, we have the folowing:
$6\int \frac{{u}^{8}}{1+{u}^{2}}du=6\left(\int {u}^{6}-{u}^{4}+{u}^{2}-1+\frac{1}{{u}^{2}+1}\right)du$
So we have the last result:
$6\left(\int {u}^{6}-{u}^{4}+{u}^{2}-1+\frac{1}{{u}^{2}+1}\right)du=\frac{6{u}^{7}}{7}-\frac{6{u}^{5}}{5}+\frac{6{u}^{3}}{3}-6u+6\mathrm{arctan}u\right)$
$\frac{6{x}^{\frac{7}{6}}}{7}-\frac{6{x}^{\frac{5}{6}}}{5}+\frac{6{x}^{\frac{1}{2}}}{3}-6{x}^{\frac{1}{6}}+6\mathrm{arctan}{x}^{\frac{1}{3}}$