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

The integral of 5/((x-4)^2) is -5/(x-4)+C

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integral of 5/((x-4)^2)

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

- \frac{5}{x - 4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Apply the Power Rule

= 5 (- \frac{1}{u})

Substitute back

u = x - 4

= 5 (- \frac{1}{x - 4})

Simplify

5 (- \frac{1}{x - 4}) : - \frac{5}{x - 4}

= - \frac{5}{x - 4}

Add a constant to the solution

= - \frac{5}{x - 4} + C