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

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

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

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

- \frac{2}{x - 1} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Apply the Power Rule

= 2 (- \frac{1}{u})

Substitute back

u = x - 1

= 2 (- \frac{1}{x - 1})

Simplify

2 (- \frac{1}{x - 1}) : - \frac{2}{x - 1}

= - \frac{2}{x - 1}

Add a constant to the solution

= - \frac{2}{x - 1} + C

cos (sin (x))

\frac{\partial}{\partial x} (e^{\frac{- 2 y}{x}})

\frac{d}{d x} (\frac{e ^{2 x}}{x ^{4}})

g (f) = f^{8}

\int \frac{1}{x + 5} d x