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

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

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

integral

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

ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{x - 1} : ln ∣ x - 1 ∣ - \frac{2}{x - 1}

= ln ∣ x - 1 ∣ - \frac{2}{x - 1}

Add a constant to the solution

= ln ∣ x - 1 ∣ - \frac{2}{x - 1} + C

sin (- u + 3 t)

\frac{1}{x + x ^{2}}

(x + 1) cos (x)

\frac{7}{x ^{2}}

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

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