What is the integral of ((3x^5))/(\sqrt[7]{x^3+1)} ?

The integral of ((3x^5))/(\sqrt[7]{x^3+1)} is 7/13 (x^3+1)^{13/7}-7/6 (x^3+1)^{6/7}+C

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integral of ((3x^5))/(\sqrt[7]{x^3+1)}

\frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{6}{7}} d u = \frac{7}{13} u^{\frac{13}{7}}

= 3 \cdot \frac{1}{3} (\frac{7}{13} u^{\frac{13}{7}} - \frac{7}{6} u^{\frac{6}{7}})

Substitute back

u = x^{3} + 1

= 3 \cdot \frac{1}{3} (\frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}})

Simplify

3 \cdot \frac{1}{3} (\frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}}) : \frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}}

= \frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}}

Add a constant to the solution

= \frac{7}{13} (x^{3} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{3} + 1)^{\frac{6}{7}} + C

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What is the integral of ((3x^5))/(\sqrt[7]{x^3+1)} ?
The integral of ((3x^5))/(\sqrt[7]{x^3+1)} is 7/13 (x^3+1)^{13/7}-7/6 (x^3+1)^{6/7}+C