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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{\partial}{\partial y} (- 2 x y)

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

sin (t) = y^{^{'}} + y

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