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

The integral of (x^3)/(\sqrt[5]{x^2+3)} is 1/2 (5/9 (x^2+3)^{9/5}-15/4 (x^2+3)^{4/5})+C

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

\int \frac{x ^{3}}{5 x ^{2} + 3} d x

\frac{1}{2} (\frac{5}{9} (x^{2} + 3)^{\frac{9}{5}} - \frac{15}{4} (x^{2} + 3)^{\frac{4}{5}}) + C

Solution steps

\int \frac{x ^{3}}{5 x ^{2} + 3} d x

Apply u-substitution

: \int \frac{u - 3}{2 5 u} d u

= \int \frac{u - 3}{2 5 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 - 3}{5 u} d u

Expand

\frac{u - 3}{5 u} : u^{\frac{4}{5}} - \frac{3}{5 u}

= \frac{1}{2} \cdot \int u^{\frac{4}{5}} - \frac{3}{5 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{4}{5}} d u - \int \frac{3}{5 u} d u)

\int u^{\frac{4}{5}} d u = \frac{5}{9} u^{\frac{9}{5}}

\int \frac{3}{5 u} d u = \frac{15}{4} u^{\frac{4}{5}}

= \frac{1}{2} (\frac{5}{9} u^{\frac{9}{5}} - \frac{15}{4} u^{\frac{4}{5}})

Substitute back

u = x^{2} + 3

= \frac{1}{2} (\frac{5}{9} (x^{2} + 3)^{\frac{9}{5}} - \frac{15}{4} (x^{2} + 3)^{\frac{4}{5}})

Add a constant to the solution

= \frac{1}{2} (\frac{5}{9} (x^{2} + 3)^{\frac{9}{5}} - \frac{15}{4} (x^{2} + 3)^{\frac{4}{5}}) + C

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What is the integral of (x^3)/(\sqrt[5]{x^2+3)} ?
The integral of (x^3)/(\sqrt[5]{x^2+3)} is 1/2 (5/9 (x^2+3)^{9/5}-15/4 (x^2+3)^{4/5})+C