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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = y - 1

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

Add a constant to the solution

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

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