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

The integral of (x^3)/(\sqrt[9]{x^2+9)} is 1/2 (9/17 (x^2+9)^{17/9}-81/8 (x^2+9)^{8/9})+C

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

\frac{1}{2} (\frac{9}{17} (x^{2} + 9)^{\frac{17}{9}} - \frac{81}{8} (x^{2} + 9)^{\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{81}{8} u^{\frac{8}{9}})

Substitute back

u = x^{2} + 9

= \frac{1}{2} (\frac{9}{17} (x^{2} + 9)^{\frac{17}{9}} - \frac{81}{8} (x^{2} + 9)^{\frac{8}{9}})

Add a constant to the solution

= \frac{1}{2} (\frac{9}{17} (x^{2} + 9)^{\frac{17}{9}} - \frac{81}{8} (x^{2} + 9)^{\frac{8}{9}}) + C

y^{^{'}} = y (3 - t y), y (0) = 1

\frac{\partial}{\partial x} (3 cos (x + y))

x \to - 100 lim (3 x^{2})

\frac{d}{d x} (2 x^{2} - 3 x)

x \to 7 lim (\frac{1}{( x - 7 ) ^{2}})

What is the integral of (x^3)/(\sqrt[9]{x^2+9)} ?
The integral of (x^3)/(\sqrt[9]{x^2+9)} is 1/2 (9/17 (x^2+9)^{17/9}-81/8 (x^2+9)^{8/9})+C