What is the integral of (8x+19)^{-2} ?

The integral of (8x+19)^{-2} is -1/(8(8x+19))+C

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integral of (8x+19)^{-2}

\int (8 x + 19)^{- 2} d x

- \frac{1}{8 ( 8 x + 19 )} + C

Solution steps

\int (8 x + 19)^{- 2} d x

Apply exponent rule:

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

= \int \frac{1}{( 8 x + 19 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{8 u ^{2}} d u

= \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}{u}

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

Substitute back

u = 8 x + 19

= \frac{1}{8} (- \frac{1}{8 x + 19})

Simplify

\frac{1}{8} (- \frac{1}{8 x + 19}) : - \frac{1}{8 ( 8 x + 19 )}

= - \frac{1}{8 ( 8 x + 19 )}

Add a constant to the solution

= - \frac{1}{8 ( 8 x + 19 )} + C

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What is the integral of (8x+19)^{-2} ?
The integral of (8x+19)^{-2} is -1/(8(8x+19))+C