What is the integral of 7\sqrt[5]{(8-2x)^8} ?

The integral of 7\sqrt[5]{(8-2x)^8} is -35/26 (8-2x)^{13/5}+C

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integral of 7\sqrt[5]{(8-2x)^8}

\int 7 5 (8 - 2 x)^{8} d x

- \frac{35}{26} (8 - 2 x)^{\frac{13}{5}} + C

Solution steps

\int 7 5 (8 - 2 x)^{8} d x

Take the constant out:

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

= 7 \cdot \int 5 (8 - 2 x)^{8} d x

Apply u-substitution

: \int - \frac{u ^{\frac{8}{5}}}{2} d u

= 7 \cdot \int - \frac{u ^{\frac{8}{5}}}{2} d u

Take the constant out:

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

= 7 (- \frac{1}{2} \cdot \int u^{\frac{8}{5}} d u)

Apply the Power Rule

: \frac{5}{13} u^{\frac{13}{5}}

= 7 (- \frac{1}{2} \cdot \frac{5}{13} u^{\frac{13}{5}})

Substitute back

u = 8 - 2 x

= 7 (- \frac{1}{2} \cdot \frac{5}{13} (8 - 2 x)^{\frac{13}{5}})

Simplify

7 (- \frac{1}{2} \cdot \frac{5}{13} (8 - 2 x)^{\frac{13}{5}}) : - \frac{35}{26} (8 - 2 x)^{\frac{13}{5}}

= - \frac{35}{26} (8 - 2 x)^{\frac{13}{5}}

Add a constant to the solution

= - \frac{35}{26} (8 - 2 x)^{\frac{13}{5}} + C

f (x) = x^{4} + 6 x^{2} - x, at x = 1

\frac{d}{d x} (ln (\frac{a}{x}))

\frac{d}{d x} (ln (x^{2} \cdot e^{x^{2} + 1}))

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

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

What is the integral of 7\sqrt[5]{(8-2x)^8} ?
The integral of 7\sqrt[5]{(8-2x)^8} is -35/26 (8-2x)^{13/5}+C