What is the integral of (2x^2+2x-7)/(x^2-4) ?

The integral of (2x^2+2x-7)/(x^2-4) is 2x-1/4 ln| x/2+1|+1/4 ln| x/2-1|+ln|x^2-4|+C

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integral of (2x^2+2x-7)/(x^2-4)

\int \frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} d x

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

Solution steps

\int \frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} d x

Expand

\frac{2 x ^{2} + 2 x - 7}{x ^{2} - 4} : \frac{2 x ^{2}}{x ^{2} - 4} + \frac{2 x}{x ^{2} - 4} - \frac{7}{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{2 x ^{2}}{x ^{2} - 4} d x + \int \frac{2 x}{x ^{2} - 4} d x - \int \frac{7}{x ^{2} - 4} d x

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

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

\int \frac{7}{x ^{2} - 4} d x = - \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

= - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + ln x^{2} - 4 - (- \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1))

Simplify

- 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2}) + ln x^{2} - 4 - (- \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)) : 2 x - \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 + ln x^{2} - 4

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

Add a constant to the solution

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

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What is the integral of (2x^2+2x-7)/(x^2-4) ?
The integral of (2x^2+2x-7)/(x^2-4) is 2x-1/4 ln| x/2+1|+1/4 ln| x/2-1|+ln|x^2-4|+C