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

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

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integral of (3x^2)/(\sqrt[3]{x^3+1)}

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

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

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

Substitute back

u = x^{3} + 1

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

Simplify

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

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

Add a constant to the solution

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

x \cdot \frac{d y}{d x} - 2 \cdot y = 3 \cdot x^{2} + 2 \cdot x

\frac{\partial}{\partial y} (arctan (y^{3} x))

\int \frac{1}{1 + x ^{2}} d x

y = sin (arccos (t))

\frac{d}{d x} (\frac{15}{x ^{2}})

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