What is the integral of (x-1)/(x^2-9) ?

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

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integral of (x-1)/(x^2-9)

\int \frac{x - 1}{x ^{2} - 9} d x

\frac{1}{2} ln x^{2} - 9 + \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Solution steps

\int \frac{x - 1}{x ^{2} - 9} d x

Expand

\frac{x - 1}{x ^{2} - 9} : \frac{x}{x ^{2} - 9} - \frac{1}{x ^{2} - 9}

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{x}{x ^{2} - 9} d x - \int \frac{1}{x ^{2} - 9} d x

\int \frac{x}{x ^{2} - 9} d x = \frac{1}{2} ln x^{2} - 9

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

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{2} \frac{3}{x ^{2}} d x

\int 4 x^{3} y^{3} d x

\int (y) d y

\int 9 tan^{3} (x) d x

y = x + 3

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