What is the integral of (-8x+4)^{-2} ?

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

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integral of (-8x+4)^{-2}

\int (- 8 x + 4)^{- 2} d x

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

Solution steps

\int (- 8 x + 4)^{- 2} d x

Apply exponent rule:

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

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

Apply u-substitution

= \int - \frac{1}{8 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - \frac{1}{8} (- \frac{1}{u})

Substitute back

u = - 8 x + 4

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

Simplify

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

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

Add a constant to the solution

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

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