What is the integral of 1/(x^{6/7)(3+x^{1/7})} ?

The integral of 1/(x^{6/7)(3+x^{1/7})} is 7ln|3+x^{1/7}|+C

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integral of 1/(x^{6/7)(3+x^{1/7})}

\int \frac{1}{x ^{\frac{6}{7}} ( 3 + x ^{\frac{1}{7}} )} d x

7 ln 3 + x^{\frac{1}{7}} + C

Solution steps

\int \frac{1}{x ^{\frac{6}{7}} ( 3 + x ^{\frac{1}{7}} )} d x

Apply u-substitution

: \int \frac{7}{u} d u

= \int \frac{7}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 ln ∣ u ∣

Substitute back

u = 3 + x^{\frac{1}{7}}

= 7 ln 3 + x^{\frac{1}{7}}

Add a constant to the solution

= 7 ln 3 + x^{\frac{1}{7}} + C

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What is the integral of 1/(x^{6/7)(3+x^{1/7})} ?
The integral of 1/(x^{6/7)(3+x^{1/7})} is 7ln|3+x^{1/7}|+C