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

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

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

\int \frac{t}{3 t ^{2} + 9} d t

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

Solution steps

\int \frac{t}{3 t ^{2} + 9} d t

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = t^{2} + 9

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

Simplify

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

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

Add a constant to the solution

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

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