What is the integral of 8/((x-4)^2) ?

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

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integral of 8/((x-4)^2)

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

- \frac{8}{x - 4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x - 4

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

Simplify

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

= - \frac{8}{x - 4}

Add a constant to the solution

= - \frac{8}{x - 4} + C

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