What is the integral of 1/(\sqrt[4]{8-3x)} ?

The integral of 1/(\sqrt[4]{8-3x)} is -4/9 (8-3x)^{3/4}+C

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integral of 1/(\sqrt[4]{8-3x)}

- \frac{4}{9} (8 - 3 x)^{\frac{3}{4}} + 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{4}{3} u^{\frac{3}{4}}

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

Substitute back

u = 8 - 3 x

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

Simplify

- \frac{1}{3} \cdot \frac{4}{3} (8 - 3 x)^{\frac{3}{4}} : - \frac{4}{9} (8 - 3 x)^{\frac{3}{4}}

= - \frac{4}{9} (8 - 3 x)^{\frac{3}{4}}

Add a constant to the solution

= - \frac{4}{9} (8 - 3 x)^{\frac{3}{4}} + C

\frac{4 x + x ^{2} - 2}{( 2 + x ) ^{2}}

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

x \to - 8 + lim (\frac{( x + 7 )}{x + 8})

\frac{d}{d x} (e^{- a ∣ x ∣})

\int \frac{2 x}{( x ^{2} + 1 ) ^{90}} d x

What is the integral of 1/(\sqrt[4]{8-3x)} ?
The integral of 1/(\sqrt[4]{8-3x)} is -4/9 (8-3x)^{3/4}+C