What is the integral of (4-2x)^{-5} ?

The integral of (4-2x)^{-5} is 1/(8(4-2x)^4)+C

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integral of (4-2x)^{-5}

\int (4 - 2 x)^{- 5} d x

\frac{1}{8 ( 4 - 2 x ) ^{4}} + C

Solution steps

\int (4 - 2 x)^{- 5} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 4 - 2 x ) ^{5}} d x

Apply u-substitution

= \int - \frac{1}{2 u ^{5}} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

= - \frac{1}{2} (- \frac{1}{4 u ^{4}})

Substitute back

u = 4 - 2 x

= - \frac{1}{2} (- \frac{1}{4 ( 4 - 2 x ) ^{4}})

Simplify

- \frac{1}{2} (- \frac{1}{4 ( 4 - 2 x ) ^{4}}) : \frac{1}{8 ( 4 - 2 x ) ^{4}}

= \frac{1}{8 ( 4 - 2 x ) ^{4}}

Add a constant to the solution

= \frac{1}{8 ( 4 - 2 x ) ^{4}} + C

What is the integral of (4-2x)^{-5} ?
The integral of (4-2x)^{-5} is 1/(8(4-2x)^4)+C