Kathleen Rausch

2021-12-26

Evaluate the indefinite integral.
$\int \frac{{x}^{6}}{{x}^{7}+2}dx$

scoollato7o

Step 1
The given integral is:
$\int \frac{{x}^{6}}{{x}^{7}+2}dx$
Let, ${x}^{7}=u$
Differentiate both sides of the above expression.
$7{x}^{6}dx=du$
${x}^{6}dx=\frac{du}{7}$
Substitute the above values into the given integral.
$\int \frac{{x}^{6}}{{x}^{7}+2}dx=\int \frac{1}{7\left(u+2\right)}du$
$=\frac{1}{7}\int \frac{1}{u+2}du$
$=\frac{1}{7}\mathrm{log}|u+2|+C$
Step 2
Substitute back $u={x}^{7}$ in the above integral.
$\int \frac{{x}^{6}}{{x}^{7}+2}dx=\frac{1}{7}\mathrm{log}|{x}^{7}+2|+C$
Therefore, the value of $\int \frac{{x}^{6}}{{x}^{7}+2}dx$ is $\frac{1}{7}\mathrm{log}|{x}^{7}+2|+C$

intacte87

$\int \frac{{x}^{6}}{{x}^{7}+2}dx$
We bring the expression $7\cdot {x}^{6}$ under the differential sign, i.e.:
$7{x}^{6}dx=d\left({x}^{7}\right),t={x}^{7}$
Then the original integral can be written as follows: We
$\int \frac{1}{7\left(t+2\right)}dt$
$\int \frac{1}{7x+14}dx$
calculate the table integral:
$\frac{1}{7}\int \frac{1}{x+2}dx=\frac{\mathrm{ln}\left(x+2\right)}{7}$
$\frac{\mathrm{ln}\left(x+2\right)}{7}+C$
or
$\mathrm{ln}\left({\left(x+2\right)}^{\frac{1}{7}}\right)+C$
To write the final answer, it remains to substitute ${x}^{7}$ for t.
$\frac{\mathrm{ln}\left({x}^{7}+2\right)}{7}+C$

Vasquez

Given:
$\int \frac{{x}^{6}}{{x}^{7}+2}dx$
$=\frac{1}{7}\int \frac{1}{u}du$
$\int \frac{1}{u}du$
This is the well-known tabular integral:
$=\mathrm{ln}\left(u\right)$
We substitute the already calculated integrals:
$\frac{1}{7}\int \frac{1}{u}du$
$=\frac{\mathrm{ln}\left(u\right)}{7}$
Reverse replacement $u={x}^{7}+2:$
$=\frac{\mathrm{ln}\left({x}^{7}+2\right)}{7}$
Solution:
$=\frac{\mathrm{ln}\left({x}^{7}+2\right)}{7}+C$

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