What is the integral of (5x^2+3x)/((x-1)(x+1)^2) ?

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

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integral of (5x^2+3x)/((x-1)(x+1)^2)

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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