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

\int tan^{3} (3 x) sec^{4} (3 x) d x

y^{^{'}} = - 0.02 y - 4

\int - cos (\frac{n x π}{2}) d x

(4 - t) (2 + t^{2})^{- 1}

x y^{^{'}} + y = x^{2} y^{2}

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