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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

Add a constant to the solution

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

(x + 2 y^{2}) d x + x y d y = 0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x, y (1) = 0, y^{^{'}} (1) = 0

\frac{d y}{d x} = \frac{x}{x ^{2} + 64}

n = 1 \sum \infty n! (2 x - 1)^{n}

\frac{d x}{d t} = - 8

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