What is the integral of (2x^2+3)^{1/3}x ?

The integral of (2x^2+3)^{1/3}x is 3/16 (2x^2+3)^{4/3}+C

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integral of (2x^2+3)^{1/3}x

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

\frac{3}{16} (2 x^{2} + 3)^{\frac{4}{3}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{v ^{\frac{1}{3}}}{2} d v

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 x^{2} + 3)^{\frac{4}{3}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} (2 x^{2} + 3)^{\frac{4}{3}} : \frac{3}{16} (2 x^{2} + 3)^{\frac{4}{3}}

= \frac{3}{16} (2 x^{2} + 3)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{16} (2 x^{2} + 3)^{\frac{4}{3}} + C

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What is the integral of (2x^2+3)^{1/3}x ?
The integral of (2x^2+3)^{1/3}x is 3/16 (2x^2+3)^{4/3}+C