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

Take the constant out:

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

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = x - 1

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

Simplify

= \frac{1}{x - 1}

Add a constant to the solution

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

y^{^{''}} + 6 y^{^{'}} + 9 = e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{1 - y ^{2}}{x}

\int \frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} d x

d er i v a t i v e y = \frac{x}{x + 1}

x \to - 9 + lim (\frac{x + 8}{x + 9})

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