What is the integral of 9/((x-3)^2) ?

The integral of 9/((x-3)^2) is -9/(x-3)+C

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integral of 9/((x-3)^2)

\int \frac{9}{( x - 3 ) ^{2}} d x

- \frac{9}{x - 3} + C

Solution steps

\int \frac{9}{( x - 3 ) ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

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

= 9 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 9 (- \frac{1}{u})

Substitute back

u = x - 3

= 9 (- \frac{1}{x - 3})

Simplify

9 (- \frac{1}{x - 3}) : - \frac{9}{x - 3}

= - \frac{9}{x - 3}

Add a constant to the solution

= - \frac{9}{x - 3} + C

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