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

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

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

\frac{1}{4} (3 v^{2} + 2)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 3 v^{2} + 2

= \frac{1}{6} \cdot \frac{3}{2} (3 v^{2} + 2)^{\frac{2}{3}}

Simplify

\frac{1}{6} \cdot \frac{3}{2} (3 v^{2} + 2)^{\frac{2}{3}} : \frac{1}{4} (3 v^{2} + 2)^{\frac{2}{3}}

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

Add a constant to the solution

= \frac{1}{4} (3 v^{2} + 2)^{\frac{2}{3}} + C

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