What is the integral of (x+1)/(x^2-4) ?

The integral of (x+1)/(x^2-4) is 1/4 ln|x+2|+3/4 ln|x-2|+C

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integral of (x+1)/(x^2-4)

\int \frac{x + 1}{x ^{2} - 4} d x

\frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x + 1}{x ^{2} - 4} d x

Take the partial fraction of

\frac{x + 1}{x ^{2} - 4} : \frac{1}{4 ( x + 2 )} + \frac{3}{4 ( x - 2 )}

= \int \frac{1}{4 ( x + 2 )} + \frac{3}{4 ( x - 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{1}{4 ( x + 2 )} d x + \int \frac{3}{4 ( x - 2 )} d x

\int \frac{1}{4 ( x + 2 )} d x = \frac{1}{4} ln ∣ x + 2 ∣

\int \frac{3}{4 ( x - 2 )} d x = \frac{3}{4} ln ∣ x - 2 ∣

= \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

f (x) = - \frac{8}{x}

x \to 3 lim (- 2 (x - 2)^{2} + 4)

- \frac{1}{2} cos (2 x)

(x^{2})^{^{'}}

\int 5 e^{- 5 x} sin (5 x) d x

What is the integral of (x+1)/(x^2-4) ?
The integral of (x+1)/(x^2-4) is 1/4 ln|x+2|+3/4 ln|x-2|+C