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

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

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

- \frac{3}{4} (1 - 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{3}{4} u^{\frac{4}{3}}

Substitute back

u = 1 - x

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

Add a constant to the solution

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

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

\frac{d}{d x} (16 sin (2 x))

\frac{1}{2 s ^{2} + 4 s + 3}

y^{^{'}} + 4 y = 3 sin (e^{4 x})

\int \frac{4 x ^{2} - 16}{x} d x

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