What is the integral of ((x(x-2))/((x-1)^3)) ?

The integral of ((x(x-2))/((x-1)^3)) is ln|x-1|+1/(2(x-1)^2)+C

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integral of ((x(x-2))/((x-1)^3))

\int (\frac{x ( x - 2 )}{( x - 1 ) ^{3}}) d x

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

Solution steps

\int \frac{x ( x - 2 )}{( x - 1 ) ^{3}} d x

Complete the square

x (x - 2) : (x - 1)^{2} - 1

= \int \frac{( x - 1 ) ^{2} - 1}{( x - 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{u ^{3}} d u

= \int \frac{u ^{2} - 1}{u ^{3}} d u

Expand

\frac{u ^{2} - 1}{u ^{3}} : \frac{1}{u} - \frac{1}{u ^{3}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= ln ∣ u ∣ - (- \frac{1}{2 u ^{2}})

Substitute back

u = x - 1

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

Simplify

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

Add a constant to the solution

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

f (x) = sinh (4 x)

\frac{t d u}{d t} = t^{2} + 3 u

\frac{\partial}{\partial x} (- 6 x^{3} + 10 y^{2} - 4 x y)

\int (- 3 x + 4) e^{- x} d x

\frac{d y}{d x} = \frac{( 2 y + 1 )}{( x + 1 )}

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