What is the integral of \sqrt[3]{8-7x^2}(-14x) ?

The integral of \sqrt[3]{8-7x^2}(-14x) is 3/4 (8-7x^2)^{4/3}+C

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integral of \sqrt[3]{8-7x^2}(-14x)

\frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}} + C

Solution steps

Simplify

Take the constant out:

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

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{3}{4} u^{\frac{4}{3}}

= - 14 (- \frac{1}{14} \cdot \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 8 - 7 x^{2}

= - 14 (- \frac{1}{14} \cdot \frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}})

Simplify

- 14 (- \frac{1}{14} \cdot \frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}}) : \frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}}

= \frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} (8 - 7 x^{2})^{\frac{4}{3}} + C

\int 37

g (x) = 3 x^{4}

f (x) = \frac{3}{x}, (5, \frac{3}{5})

\int x^{4} e^{7 x} d x

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

What is the integral of \sqrt[3]{8-7x^2}(-14x) ?
The integral of \sqrt[3]{8-7x^2}(-14x) is 3/4 (8-7x^2)^{4/3}+C