What is the integral of (2x-6)/(x^2-81) ?

The integral of (2x-6)/(x^2-81) is ln|x^2-81|+1/3 (ln| x/9+1|-ln| x/9-1|)+C

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integral of (2x-6)/(x^2-81)

\int \frac{2 x - 6}{x ^{2} - 81} d x

ln x^{2} - 81 + \frac{1}{3} (ln \frac{x}{9} + 1 - ln \frac{x}{9} - 1) + C

Solution steps

\int \frac{2 x - 6}{x ^{2} - 81} d x

Expand

\frac{2 x - 6}{x ^{2} - 81} : \frac{2 x}{x ^{2} - 81} - \frac{6}{x ^{2} - 81}

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}{x ^{2} - 81} d x - \int \frac{6}{x ^{2} - 81} d x

\int \frac{2 x}{x ^{2} - 81} d x = ln x^{2} - 81

\int \frac{6}{x ^{2} - 81} d x = - \frac{1}{3} (ln \frac{x}{9} + 1 - ln \frac{x}{9} - 1)

= ln x^{2} - 81 - (- \frac{1}{3} (ln \frac{x}{9} + 1 - ln \frac{x}{9} - 1))

Simplify

= ln x^{2} - 81 + \frac{1}{3} (ln \frac{x}{9} + 1 - ln \frac{x}{9} - 1)

Add a constant to the solution

= ln x^{2} - 81 + \frac{1}{3} (ln \frac{x}{9} + 1 - ln \frac{x}{9} - 1) + C

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\int sin^{2} (\frac{π x}{a}) d x

What is the integral of (2x-6)/(x^2-81) ?
The integral of (2x-6)/(x^2-81) is ln|x^2-81|+1/3 (ln| x/9+1|-ln| x/9-1|)+C