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}

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

4 x^{3} - 3 x^{2} + 4, at x = 2

\frac{d y}{d x} = 2 y - x^{2}

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

\frac{\partial}{\partial x} (\frac{5}{e ^{x}})

y^{^{'}} + 4 y = 3 sin (2 t)

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