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

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

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

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

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

Solution steps

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

Expand

\frac{x + 2}{x ^{2} - 4} : \frac{x}{x ^{2} - 4} + \frac{2}{x ^{2} - 4}

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

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

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

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

Add a constant to the solution

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

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