What is the integral of (x^3)/(\sqrt[4]{x^2+9)} ?

The integral of (x^3)/(\sqrt[4]{x^2+9)} is 2/7 (x^2-12)(x^2+9)^{3/4}+C

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integral of (x^3)/(\sqrt[4]{x^2+9)}

\int \frac{x ^{3}}{4 x ^{2} + 9} d x

\frac{2}{7} (x^{2} - 12) (x^{2} + 9)^{\frac{3}{4}} + C

Solution steps

\int \frac{x ^{3}}{4 x ^{2} + 9} d x

Apply u-substitution

= \int \frac{u - 9}{2 4 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 - 9}{4 u} d u

Expand

\frac{u - 9}{4 u} : u^{\frac{3}{4}} - \frac{9}{4 u}

= \frac{1}{2} \cdot \int u^{\frac{3}{4}} - \frac{9}{4 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{3}{4}} d u - \int \frac{9}{4 u} d u)

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{9}{4 u} d u = 12 u^{\frac{3}{4}}

= \frac{1}{2} (\frac{4}{7} u^{\frac{7}{4}} - 12 u^{\frac{3}{4}})

Substitute back

u = x^{2} + 9

= \frac{1}{2} (\frac{4}{7} (x^{2} + 9)^{\frac{7}{4}} - 12 (x^{2} + 9)^{\frac{3}{4}})

Simplify

\frac{1}{2} (\frac{4}{7} (x^{2} + 9)^{\frac{7}{4}} - 12 (x^{2} + 9)^{\frac{3}{4}}) : \frac{2}{7} (x^{2} - 12) (x^{2} + 9)^{\frac{3}{4}}

= \frac{2}{7} (x^{2} - 12) (x^{2} + 9)^{\frac{3}{4}}

Add a constant to the solution

= \frac{2}{7} (x^{2} - 12) (x^{2} + 9)^{\frac{3}{4}} + C

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What is the integral of (x^3)/(\sqrt[4]{x^2+9)} ?
The integral of (x^3)/(\sqrt[4]{x^2+9)} is 2/7 (x^2-12)(x^2+9)^{3/4}+C