What is the integral of \sqrt[4]{8-3x} ?

The integral of \sqrt[4]{8-3x} is -4/15 (8-3x)^{5/4}+C

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integral of \sqrt[4]{8-3x}

\int 4 8 - 3 x d x

- \frac{4}{15} (8 - 3 x)^{\frac{5}{4}} + C

Solution steps

\int 4 8 - 3 x d x

Apply u-substitution

: \int - \frac{4 u}{3} d u

= \int - \frac{4 u}{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 4 u d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= - \frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 8 - 3 x

= - \frac{1}{3} \cdot \frac{4}{5} (8 - 3 x)^{\frac{5}{4}}

Simplify

- \frac{1}{3} \cdot \frac{4}{5} (8 - 3 x)^{\frac{5}{4}} : - \frac{4}{15} (8 - 3 x)^{\frac{5}{4}}

= - \frac{4}{15} (8 - 3 x)^{\frac{5}{4}}

Add a constant to the solution

= - \frac{4}{15} (8 - 3 x)^{\frac{5}{4}} + C

12 arctan (s)

\int_{1}^{2} \frac{2}{( x - 5 ) ^{3}} d x

\int 4^{x} - \frac{1}{3 x} d x

8 (ln (y)) y^{^{'}} - x^{9} y = 0

\int \frac{3 x ^{5} + 2}{- 5 x ^{5} + 5 x} d x

What is the integral of \sqrt[4]{8-3x} ?
The integral of \sqrt[4]{8-3x} is -4/15 (8-3x)^{5/4}+C