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

y^{^{'}} = x y^{2} + 2 y^{2}, y (0) = 1

d er i v a t i v e f (x) = \frac{2 x ^{2} - 6 x + 5}{2 ( 2 x - 3 )}

\frac{d y}{d x} = sec^{2} (x) (2 + y)^{2}

in v erse l a pl a ce \frac{s + 6}{s ^{2} + 6 s + 25}

\int \frac{t - 3}{t + 1} d t

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