What is the integral of (x^3)/(\sqrt[9]{x^2+7)} ?

The integral of (x^3)/(\sqrt[9]{x^2+7)} is 1/2 (9/17 (x^2+7)^{17/9}-63/8 (x^2+7)^{8/9})+C

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integral of (x^3)/(\sqrt[9]{x^2+7)}

\int \frac{x ^{3}}{9 x ^{2} + 7} d x

\frac{1}{2} (\frac{9}{17} (x^{2} + 7)^{\frac{17}{9}} - \frac{63}{8} (x^{2} + 7)^{\frac{8}{9}}) + C

Solution steps

\int \frac{x ^{3}}{9 x ^{2} + 7} d x

Apply u-substitution

: \int \frac{u - 7}{2 9 u} d u

= \int \frac{u - 7}{2 9 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 - 7}{9 u} d u

Expand

\frac{u - 7}{9 u} : u^{\frac{8}{9}} - \frac{7}{9 u}

= \frac{1}{2} \cdot \int u^{\frac{8}{9}} - \frac{7}{9 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{8}{9}} d u - \int \frac{7}{9 u} d u)

\int u^{\frac{8}{9}} d u = \frac{9}{17} u^{\frac{17}{9}}

\int \frac{7}{9 u} d u = \frac{63}{8} u^{\frac{8}{9}}

= \frac{1}{2} (\frac{9}{17} u^{\frac{17}{9}} - \frac{63}{8} u^{\frac{8}{9}})

Substitute back

u = x^{2} + 7

= \frac{1}{2} (\frac{9}{17} (x^{2} + 7)^{\frac{17}{9}} - \frac{63}{8} (x^{2} + 7)^{\frac{8}{9}})

Add a constant to the solution

= \frac{1}{2} (\frac{9}{17} (x^{2} + 7)^{\frac{17}{9}} - \frac{63}{8} (x^{2} + 7)^{\frac{8}{9}}) + C

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What is the integral of (x^3)/(\sqrt[9]{x^2+7)} ?
The integral of (x^3)/(\sqrt[9]{x^2+7)} is 1/2 (9/17 (x^2+7)^{17/9}-63/8 (x^2+7)^{8/9})+C