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

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

Popular >

integral of 1/(\sqrt[3]{1-3x)}

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

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

Substitute back

u = 1 - 3 x

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

Simplify

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

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

Add a constant to the solution

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

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

\int (x^{3} - 5 x^{2} + 2 x + 4) sin (4 x) d x

\int (2 - x)^{6} d x

\int x \frac{1}{3} d x

\int \frac{e ^{x - 1}}{x - 1} d x

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