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)}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

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

Apply the Power Rule

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

= 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

f (x) = e^{- 4 x} cos (- 4 x)

\int (x \cdot 9^{2 x^{2} - 8}) d x

\frac{\partial}{\partial x} (1 + x ln (x y - 9))

\frac{2 s + 2}{s ^{2} + 2 s + 5}

\frac{\partial}{\partial y} (\frac{x}{1 + y ^{2}})

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