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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{x} + ln (x))

y^{^{''}} - 4 y^{^{'}} = 6 t

\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}

\int \frac{3}{x ( x ^{4} + 1 )} d x

\frac{\partial}{\partial z} (cos (x y))

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