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

The integral of ((x+1)^2)/((x-1)^2) is x+4(ln|x-1|-1/(x-1))+C

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

integral

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

x + 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + C

Solution steps

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

Long division

\frac{( x + 1 ) ^{2}}{( x - 1 ) ^{2}} : 1 + \frac{4 x}{x ^{2} - 2 x + 1}

= \int 1 + \frac{4 x}{x ^{2} - 2 x + 1} d x

Apply the Sum Rule:

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

= \int 1 d x + \int \frac{4 x}{x ^{2} - 2 x + 1} d x

\int 1 d x = x

\int \frac{4 x}{x ^{2} - 2 x + 1} d x = 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

= x + 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

Add a constant to the solution

= x + 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + C

tan (2 - 3 x)

e^{x^{2}} x^{2}

3 x^{2} - 9 x + 5

\frac{e ^{x}}{3}

\frac{1}{5 x}

What is the integral of ((x+1)^2)/((x-1)^2) ?
The integral of ((x+1)^2)/((x-1)^2) is x+4(ln|x-1|-1/(x-1))+C