What is the integral of (x^4)/(\sqrt[7]{x^5+1)} ?

The integral of (x^4)/(\sqrt[7]{x^5+1)} is 7/30 (x^5+1)^{6/7}+C

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integral of (x^4)/(\sqrt[7]{x^5+1)}

\int \frac{x ^{4}}{7 x ^{5} + 1} d x

\frac{7}{30} (x^{5} + 1)^{\frac{6}{7}} + C

Solution steps

\int \frac{x ^{4}}{7 x ^{5} + 1} d x

Apply u-substitution

= \int \frac{1}{5 7 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{7 u} d u

Apply the Power Rule

= \frac{1}{5} \cdot \frac{7}{6} u^{\frac{6}{7}}

Substitute back

u = x^{5} + 1

= \frac{1}{5} \cdot \frac{7}{6} (x^{5} + 1)^{\frac{6}{7}}

Simplify

\frac{1}{5} \cdot \frac{7}{6} (x^{5} + 1)^{\frac{6}{7}} : \frac{7}{30} (x^{5} + 1)^{\frac{6}{7}}

= \frac{7}{30} (x^{5} + 1)^{\frac{6}{7}}

Add a constant to the solution

= \frac{7}{30} (x^{5} + 1)^{\frac{6}{7}} + C

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What is the integral of (x^4)/(\sqrt[7]{x^5+1)} ?
The integral of (x^4)/(\sqrt[7]{x^5+1)} is 7/30 (x^5+1)^{6/7}+C