sagnuhh

2020-10-25

Solve. $\int \frac{\sqrt{x}}{1+\sqrt[3]{x}}dx=?$

casincal

Skilled2020-10-26Added 82 answers

Notice that if $u={x}^{\frac{1}{6}},then\text{}{u}^{2}={x}^{\frac{1}{3}}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{u}^{3}={x}^{\frac{1}{2}}$ .

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:

$6\int \frac{{u}^{8}}{1+{u}^{2}}du=6(\int {u}^{6}-{u}^{4}+{u}^{2}-1+\frac{1}{{u}^{2}+1})du$

So we have the last result:

$6(\int {u}^{6}-{u}^{4}+{u}^{2}-1+\frac{1}{{u}^{2}+1})du=\frac{6{u}^{7}}{7}-\frac{6{u}^{5}}{5}+\frac{6{u}^{3}}{3}-6u+6\mathrm{arctan}u)$

$\frac{6{x}^{\frac{7}{6}}}{7}-\frac{6{x}^{\frac{5}{6}}}{5}+\frac{6{x}^{\frac{1}{2}}}{3}-6{x}^{\frac{1}{6}}+6\mathrm{arctan}{x}^{\frac{1}{3}}$

Also we have that

So that,

From here, we can simply use some polynomial division to simplify the quotient.Notice then that:

Thus, we have the folowing:

So we have the last result:

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