What is the integral of (y-1)^{-2} ?

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

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integral of (y-1)^{-2}

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

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

Solution steps

\int (y - 1)^{- 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 - 1

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

Add a constant to the solution

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

\int x (x^{2} + 3)^{3} d x

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\int (1 - x)^{9} d x

f (x) = \frac{sin ( x )}{cos ( x )}

\int x^{2} - 2 d x