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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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