What is the integral of 9/((t-8)^2) ?

The integral of 9/((t-8)^2) is -9/(t-8)+C

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integral of 9/((t-8)^2)

\int \frac{9}{( t - 8 ) ^{2}} d t

- \frac{9}{t - 8} + C

Solution steps

\int \frac{9}{( t - 8 ) ^{2}} d t

Take the constant out:

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

= 9 \cdot \int \frac{1}{( t - 8 ) ^{2}} d t

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 = t - 8

= 9 (- \frac{1}{t - 8})

Simplify

9 (- \frac{1}{t - 8}) : - \frac{9}{t - 8}

= - \frac{9}{t - 8}

Add a constant to the solution

= - \frac{9}{t - 8} + C

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