What is the integral of (x^3)/(\sqrt[9]{x^2+5)} ?

The integral of (x^3)/(\sqrt[9]{x^2+5)} is 1/2 (9/17 (x^2+5)^{17/9}-45/8 (x^2+5)^{8/9})+C

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integral of (x^3)/(\sqrt[9]{x^2+5)}

\frac{1}{2} (\frac{9}{17} (x^{2} + 5)^{\frac{17}{9}} - \frac{45}{8} (x^{2} + 5)^{\frac{8}{9}}) + 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{8}{9}} d u = \frac{9}{17} u^{\frac{17}{9}}

= \frac{1}{2} (\frac{9}{17} u^{\frac{17}{9}} - \frac{45}{8} u^{\frac{8}{9}})

Substitute back

u = x^{2} + 5

= \frac{1}{2} (\frac{9}{17} (x^{2} + 5)^{\frac{17}{9}} - \frac{45}{8} (x^{2} + 5)^{\frac{8}{9}})

Add a constant to the solution

= \frac{1}{2} (\frac{9}{17} (x^{2} + 5)^{\frac{17}{9}} - \frac{45}{8} (x^{2} + 5)^{\frac{8}{9}}) + C

\frac{d}{d x} (\frac{x}{∣ x ∣})

\frac{\partial}{\partial x} (π)

- x \frac{d y}{d x} + y = 0

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

\int 24 x^{2} + 16 x d x

What is the integral of (x^3)/(\sqrt[9]{x^2+5)} ?
The integral of (x^3)/(\sqrt[9]{x^2+5)} is 1/2 (9/17 (x^2+5)^{17/9}-45/8 (x^2+5)^{8/9})+C