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

The integral of (3x+1)(3x^2+2x)^{1/3} is 3/8 (3x^2+2x)^{4/3}+C

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integral of (3x+1)(3x^2+2x)^{1/3}

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 3 x^{2} + 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{x ^{2} - 2 x + 2})

\frac{df}{d x} f (x) = cos (2 x)

(cos (7 x)) y^{^{'}} - 7 (tan (7 x)) y = 9 sec (7 x)

\int_{1}^{2} \frac{3}{x ^{4}} d x

\frac{d r}{d θ} = \frac{r ^{2} + 4 r θ}{2 θ ^{2}}

What is the integral of (3x+1)(3x^2+2x)^{1/3} ?
The integral of (3x+1)(3x^2+2x)^{1/3} is 3/8 (3x^2+2x)^{4/3}+C