What is the integral of (x^3)/(\sqrt[7]{x^2+1)} ?

The integral of (x^3)/(\sqrt[7]{x^2+1)} is 1/2 (7/13 (x^2+1)^{13/7}-7/6 (x^2+1)^{6/7})+C

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integral of (x^3)/(\sqrt[7]{x^2+1)}

\int \frac{x ^{3}}{7 x ^{2} + 1} d x

\frac{1}{2} (\frac{7}{13} (x^{2} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{2} + 1)^{\frac{6}{7}}) + C

Solution steps

\int \frac{x ^{3}}{7 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{u - 1}{2 7 u} d u

= \int \frac{u - 1}{2 7 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u - 1}{7 u} d u

Expand

\frac{u - 1}{7 u} : u^{\frac{6}{7}} - \frac{1}{7 u}

= \frac{1}{2} \cdot \int u^{\frac{6}{7}} - \frac{1}{7 u} d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int u^{\frac{6}{7}} d u - \int \frac{1}{7 u} d u)

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

\int \frac{1}{7 u} d u = \frac{7}{6} u^{\frac{6}{7}}

= \frac{1}{2} (\frac{7}{13} u^{\frac{13}{7}} - \frac{7}{6} u^{\frac{6}{7}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (\frac{7}{13} (x^{2} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{2} + 1)^{\frac{6}{7}})

Add a constant to the solution

= \frac{1}{2} (\frac{7}{13} (x^{2} + 1)^{\frac{13}{7}} - \frac{7}{6} (x^{2} + 1)^{\frac{6}{7}}) + C

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What is the integral of (x^3)/(\sqrt[7]{x^2+1)} ?
The integral of (x^3)/(\sqrt[7]{x^2+1)} is 1/2 (7/13 (x^2+1)^{13/7}-7/6 (x^2+1)^{6/7})+C