What is the integral of (5x-2)/(x^2-4) ?

The integral of (5x-2)/(x^2-4) is 5/2 ln|x^2-4|+1/2 ln| x/2+1|-1/2 ln| x/2-1|+C

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integral of (5x-2)/(x^2-4)

\int \frac{5 x - 2}{x ^{2} - 4} d x

\frac{5}{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{5 x - 2}{x ^{2} - 4} d x

Expand

\frac{5 x - 2}{x ^{2} - 4} : \frac{5 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{5 x}{x ^{2} - 4} d x - \int \frac{2}{x ^{2} - 4} d x

\int \frac{5 x}{x ^{2} - 4} d x = \frac{5}{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{5}{2} ln x^{2} - 4 - (- \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1)

Simplify

= \frac{5}{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{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

\frac{d}{d x} (sin (8 x + 4))

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

\frac{d}{d x} (cos (a t))

\frac{9}{x ^{2}}

- 0.025 x^{2} + 8 x

What is the integral of (5x-2)/(x^2-4) ?
The integral of (5x-2)/(x^2-4) is 5/2 ln|x^2-4|+1/2 ln| x/2+1|-1/2 ln| x/2-1|+C