What is the integral of (x^2+11x)/((x-1)(x+1)^2) ?

The integral of (x^2+11x)/((x-1)(x+1)^2) is 3ln|x-1|-2ln|x+1|-5/(x+1)+C

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integral of (x^2+11x)/((x-1)(x+1)^2)

\int \frac{x ^{2} + 11 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

3 ln ∣ x - 1 ∣ - 2 ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

Solution steps

\int \frac{x ^{2} + 11 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 11 x}{( x - 1 ) ( x + 1 ) ^{2}} : \frac{3}{x - 1} - \frac{2}{x + 1} + \frac{5}{( x + 1 ) ^{2}}

= \int \frac{3}{x - 1} - \frac{2}{x + 1} + \frac{5}{( x + 1 ) ^{2}} 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{3}{x - 1} d x - \int \frac{2}{x + 1} d x + \int \frac{5}{( x + 1 ) ^{2}} d x

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

\int \frac{5}{( x + 1 ) ^{2}} d x = - \frac{5}{x + 1}

= 3 ln ∣ x - 1 ∣ - 2 ln ∣ x + 1 ∣ - \frac{5}{x + 1}

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ - 2 ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

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What is the integral of (x^2+11x)/((x-1)(x+1)^2) ?
The integral of (x^2+11x)/((x-1)(x+1)^2) is 3ln|x-1|-2ln|x+1|-5/(x+1)+C