What is the integral of 8/((y-9)^3) ?

The integral of 8/((y-9)^3) is -4/((y-9)^2)+C

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integral of 8/((y-9)^3)

\int \frac{8}{( y - 9 ) ^{3}} d y

- \frac{4}{( y - 9 ) ^{2}} + C

Solution steps

\int \frac{8}{( y - 9 ) ^{3}} d y

Take the constant out:

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

= 8 \cdot \int \frac{1}{( y - 9 ) ^{3}} d y

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= 8 \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 8 (- \frac{1}{2 u ^{2}})

Substitute back

u = y - 9

= 8 (- \frac{1}{2 ( y - 9 ) ^{2}})

Simplify

8 (- \frac{1}{2 ( y - 9 ) ^{2}}) : - \frac{4}{( y - 9 ) ^{2}}

= - \frac{4}{( y - 9 ) ^{2}}

Add a constant to the solution

= - \frac{4}{( y - 9 ) ^{2}} + C

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What is the integral of 8/((y-9)^3) ?
The integral of 8/((y-9)^3) is -4/((y-9)^2)+C