What is the integral of (x-1)^{-3} ?

The integral of (x-1)^{-3} is -1/(2(x-1)^2)+C

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integral of (x-1)^{-3}

\int (x - 1)^{- 3} d x

- \frac{1}{2 ( x - 1 ) ^{2}} + C

Solution steps

\int (x - 1)^{- 3} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( x - 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = x - 1

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

Add a constant to the solution

= - \frac{1}{2 ( x - 1 ) ^{2}} + C

e^{x}, e^{- x}, [- 2, 1]

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

\frac{\partial}{\partial x} (y^{6} cos (6 x))

\int z sin (8 z) d z

\frac{\partial}{\partial x} (x e^{x} cos (y))

What is the integral of (x-1)^{-3} ?
The integral of (x-1)^{-3} is -1/(2(x-1)^2)+C