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

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

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

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

2 ln ∣ x - 1 ∣ + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= 2 ln ∣ x - 1 ∣ + x

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + x + C

\frac{d}{d x} (12 x^{3} + 12 x^{2})

\int \frac{x - x ^{3} e ^{x} + x ^{2}}{x ^{3}} d x

x \to 0 lim (x cos (\frac{11}{2} x))

n \to \infty lim (\frac{1}{n})

\frac{d}{d x} (ln (9 - 2 x^{2}))

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