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

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

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

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

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

Substitute back

u = 3 x - 8

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

Simplify

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

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

Add a constant to the solution

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

\int 144 - x^{2} d x

y = e^{x}, y = 10 e^{- x} + 3, x = 0

x \to 1 lim (x^{4})

\frac{8}{x ^{2}}

f (x) = x^{2} - 7, (5, 18)

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