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

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

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

\int \frac{x + 3}{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 ∣) + C

Solution steps

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

Expand

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (18 x e^{x^{2}})

t^{2} y^{^{''}} - 2 y = 3 t^{2} - 1, y (1) = \frac{3}{2}, y^{^{'}} (1) = 3

\frac{s}{( s + 4 ) ^{2} ( s ^{2} - 9 )}

\int \frac{9 e ^{- \frac{9}{8} x}}{32} x d x

\int 3 e^{x} - sin (x) d x

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