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

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

Popular >

integral of \sqrt[3]{3x+2}

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

Solution steps

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}{4} u^{\frac{4}{3}}

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

Substitute back

u = 3 x + 2

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 16 x ^{2} + 9 ) ^{2}} d x

\frac{7}{x}, at x = 6

\frac{\partial}{\partial x} (sin (- x) cos (8 y))

2 tan (cos (x))

\int \frac{cos ( y )}{sin ( y )} d y

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