What is the integral of 1/((2-y)^3) ?

The integral of 1/((2-y)^3) is 1/(2(2-y)^2)+C

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integral of 1/((2-y)^3)

\int \frac{1}{( 2 - y ) ^{3}} d y

\frac{1}{2 ( 2 - y ) ^{2}} + C

Solution steps

\int \frac{1}{( 2 - y ) ^{3}} d y

Apply u-substitution

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

= \int - \frac{1}{u ^{3}} d u

Take the constant out:

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

= - \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 - y

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

Simplify

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

Add a constant to the solution

= \frac{1}{2 ( 2 - y ) ^{2}} + C

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What is the integral of 1/((2-y)^3) ?
The integral of 1/((2-y)^3) is 1/(2(2-y)^2)+C