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

The integral of (3x+1)\sqrt[3]{3x^2+2x} is 1/(8\sqrt[3]{3)}(9x^2+6x)^{4/3}+C

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

\int (3 x + 1) 3 3 x^{2} + 2 x d x

\frac{1}{8 3 3} (9 x^{2} + 6 x)^{\frac{4}{3}} + C

Solution steps

\int (3 x + 1) 3 3 x^{2} + 2 x d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{3 v}{2} d v

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

= \frac{1}{3 3 3} \cdot \frac{1}{2} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= \frac{1}{3 3 3} \cdot \frac{1}{2} \cdot \frac{3}{4} ((3 x + 1)^{2} - 1)^{\frac{4}{3}}

Simplify

\frac{1}{3 3 3} \cdot \frac{1}{2} \cdot \frac{3}{4} ((3 x + 1)^{2} - 1)^{\frac{4}{3}} : \frac{1}{8 3 3} (9 x^{2} + 6 x)^{\frac{4}{3}}

= \frac{1}{8 3 3} (9 x^{2} + 6 x)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{8 3 3} (9 x^{2} + 6 x)^{\frac{4}{3}} + C

\int \frac{x ^{3} + x ^{2} + x - 2}{x ^{3} - x ^{2}} d x

4 - x^{2}, \frac{1}{4} x^{3} - 2, 1, 2

\frac{d y}{d x} = \frac{8 x ^{2}}{9 + x ^{3}}

h^{2} + 7 h + 40

y^{^{'}} = x y^{8}

What is the integral of (3x+1)\sqrt[3]{3x^2+2x} ?
The integral of (3x+1)\sqrt[3]{3x^2+2x} is 1/(8\sqrt[3]{3)}(9x^2+6x)^{4/3}+C