What is the integral of (x^3)/(\sqrt[7]{x^2+5)} ?

The integral of (x^3)/(\sqrt[7]{x^2+5)} is 1/2 (7/13 (x^2+5)^{13/7}-35/6 (x^2+5)^{6/7})+C

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integral of (x^3)/(\sqrt[7]{x^2+5)}

\frac{1}{2} (\frac{7}{13} (x^{2} + 5)^{\frac{13}{7}} - \frac{35}{6} (x^{2} + 5)^{\frac{6}{7}}) + C

Solution steps

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}}

= \frac{1}{2} (\frac{7}{13} u^{\frac{13}{7}} - \frac{35}{6} u^{\frac{6}{7}})

Substitute back

u = x^{2} + 5

= \frac{1}{2} (\frac{7}{13} (x^{2} + 5)^{\frac{13}{7}} - \frac{35}{6} (x^{2} + 5)^{\frac{6}{7}})

Add a constant to the solution

= \frac{1}{2} (\frac{7}{13} (x^{2} + 5)^{\frac{13}{7}} - \frac{35}{6} (x^{2} + 5)^{\frac{6}{7}}) + C

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x \to - \infty lim (- \frac{1}{x})

\int 2 sin^{3} (\frac{x}{2}) d x

x^{2}, [0, 1]

What is the integral of (x^3)/(\sqrt[7]{x^2+5)} ?
The integral of (x^3)/(\sqrt[7]{x^2+5)} is 1/2 (7/13 (x^2+5)^{13/7}-35/6 (x^2+5)^{6/7})+C