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

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

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integral of 1/((x-1)^2)

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

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

Solution steps

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

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 = x - 1

= - \frac{1}{x - 1}

Add a constant to the solution

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

\frac{\partial}{\partial x} (5 x y)

\frac{d}{d x} ((e^{- x} - e^{5}) (e^{x} + e^{- 3}))

\frac{d}{d x} (2^{2 x + 1})

y = \frac{1}{x - 2}, at x = 4

y = 5 - 3 x^{2}