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

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

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

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

- \frac{1}{y - 3} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = y - 3

= - \frac{1}{y - 3}

Add a constant to the solution

= - \frac{1}{y - 3} + C

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