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

The integral of (x^3)/(\sqrt[3]{x^2+3)} is 1/20 (6(x^2+3)^{5/3}-45(x^2+3)^{2/3})+C

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

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

\frac{1}{20} (6 (x^{2} + 3)^{\frac{5}{3}} - 45 (x^{2} + 3)^{\frac{2}{3}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{3 3 u ^{\frac{2}{3}} + 3} d u

= \int \frac{u ^{\frac{1}{3}}}{3 3 u ^{\frac{2}{3}} + 3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{3}}}{3 u ^{\frac{2}{3}} + 3} d u

Apply u-substitution

: \int \frac{3 ( v - 3 )}{2 3 v} d v

= \frac{1}{3} \cdot \int \frac{3 ( v - 3 )}{2 3 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int \frac{v - 3}{3 v} d v

Expand

\frac{v - 3}{3 v} : v^{\frac{2}{3}} - \frac{3}{3 v}

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int v^{\frac{2}{3}} - \frac{3}{3 v} d v

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}{3} \cdot \frac{3}{2} (\int v^{\frac{2}{3}} d v - \int \frac{3}{3 v} d v)

\int v^{\frac{2}{3}} d v = \frac{3}{5} v^{\frac{5}{3}}

\int \frac{3}{3 v} d v = \frac{9}{2} v^{\frac{2}{3}}

= \frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} v^{\frac{5}{3}} - \frac{9}{2} v^{\frac{2}{3}})

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} ((x^{3})^{\frac{2}{3}} + 3)^{\frac{5}{3}} - \frac{9}{2} ((x^{3})^{\frac{2}{3}} + 3)^{\frac{2}{3}}) : \frac{1}{20} (6 (x^{2} + 3)^{\frac{5}{3}} - 45 (x^{2} + 3)^{\frac{2}{3}})

= \frac{1}{20} (6 (x^{2} + 3)^{\frac{5}{3}} - 45 (x^{2} + 3)^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{20} (6 (x^{2} + 3)^{\frac{5}{3}} - 45 (x^{2} + 3)^{\frac{2}{3}}) + C

Y^{3}

\int \frac{y}{4} d y

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\int \frac{( 2 x ^{2} + 1 )}{x} d x

\frac{d}{d x} (sech^{2} (x - 1))

What is the integral of (x^3)/(\sqrt[3]{x^2+3)} ?
The integral of (x^3)/(\sqrt[3]{x^2+3)} is 1/20 (6(x^2+3)^{5/3}-45(x^2+3)^{2/3})+C