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

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

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

\frac{9}{4} (t^{2} + 3)^{\frac{2}{3}} + C

Solution steps

Take the constant out:

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

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

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

Substitute back

u = t^{2} + 3

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

Simplify

3 \cdot \frac{1}{2} \cdot \frac{3}{2} (t^{2} + 3)^{\frac{2}{3}} : \frac{9}{4} (t^{2} + 3)^{\frac{2}{3}}

= \frac{9}{4} (t^{2} + 3)^{\frac{2}{3}}

Add a constant to the solution

= \frac{9}{4} (t^{2} + 3)^{\frac{2}{3}} + C

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