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

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

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

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

ln ∣ 1 - x ∣ + \frac{1}{1 - x} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - (- ln ∣ u ∣ - \frac{1}{u})

Substitute back

u = 1 - x

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

Simplify

= ln ∣ 1 - x ∣ + \frac{1}{1 - x}

Add a constant to the solution

= ln ∣ 1 - x ∣ + \frac{1}{1 - x} + C

x \to 0 lim (\frac{9 x ^{2} - 8 x}{x})

- y + y^{^{'}} = x + e^{2} x

y^{^{'}} + \frac{2}{x} y = x^{2} y^{2}

\frac{d}{d x} (\frac{cos ( x )}{x})

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