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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

x \to - 1 - lim (\frac{x - 2}{x ^{2} - 1})

x \to + 0 lim (x)

e^{2 x}

\int (2 x + 3)^{4} d x

\int e^{uv} d u

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