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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{5 v}{v ^{2} - 4} d v = \frac{5}{2} ln v^{2} - 4

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

= \frac{5}{2} ln v^{2} - 4 - (- \frac{1}{2} ln \frac{v}{2} + 1 + \frac{1}{2} ln \frac{v}{2} - 1)

Simplify

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

Add a constant to the solution

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

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