What is the integral of (1-3x)^{1/3} ?

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

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integral of (1-3x)^{1/3}

\int (1 - 3 x)^{\frac{1}{3}} d x

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

Solution steps

\int (1 - 3 x)^{\frac{1}{3}} d x

Apply u-substitution

: \int - \frac{u ^{\frac{1}{3}}}{3} d u

= \int - \frac{u ^{\frac{1}{3}}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{3} \cdot \int u^{\frac{1}{3}} d u

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

x \to π lim (tan (\frac{2}{3} x))

f (x) = \frac{x ^{2}}{2 + x}

n \to \infty lim (\frac{n}{1})

\int e^{3 x} cos (4 x) d x

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

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