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

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

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

- \frac{1}{4} (1 - 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 = 1 - 3 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x + 1) - x + 1)

x \to 5 lim (4 x^{2} - 8)

\frac{d}{d x} (x x + 9)

\frac{9}{4} t^{2} e^{5 t}

\int \frac{4 x - 1}{x ^{2} + x - 2} d x

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