What is the integral of (x^3)/(\sqrt[5]{x^2+1)} ?

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

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integral of (x^3)/(\sqrt[5]{x^2+1)}

\int \frac{x ^{3}}{5 x ^{2} + 1} d x

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

Solution steps

\int \frac{x ^{3}}{5 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{u - 1}{2 5 u} d u

= \int \frac{u - 1}{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 - 1}{5 u} d u

Expand

\frac{u - 1}{5 u} : u^{\frac{4}{5}} - \frac{1}{5 u}

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

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

\int \frac{1}{5 u} d u = \frac{5}{4} u^{\frac{4}{5}}

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

Substitute back

u = x^{2} + 1

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

Add a constant to the solution

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

\frac{d}{d x} ((arctan (x))^{x})

\int 36 x^{5} d x

\frac{\partial}{\partial x} (4 e^{x} ln (y))

x \to 3 lim (18 - x^{2})

\frac{d}{d x} (arctan (x^{2} y))

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