What is the integral of (x^2-1)/(x^2-16) ?

The integral of (x^2-1)/(x^2-16) is -15/8 (ln| x/4+1|-ln| x/4-1|)+x+C

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integral of (x^2-1)/(x^2-16)

\int \frac{x ^{2} - 1}{x ^{2} - 16} d x

- \frac{15}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + x + C

Solution steps

\int \frac{x ^{2} - 1}{x ^{2} - 16} d x

\frac{x ^{2} - 1}{x ^{2} - 16} = \frac{15}{x ^{2} - 16} + 1

= \int \frac{15}{x ^{2} - 16} + 1 d x

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{15}{x ^{2} - 16} d x + \int 1 d x

\int \frac{15}{x ^{2} - 16} d x = - \frac{15}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

\int 1 d x = x

= - \frac{15}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + x

Add a constant to the solution

= - \frac{15}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + x + C

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What is the integral of (x^2-1)/(x^2-16) ?
The integral of (x^2-1)/(x^2-16) is -15/8 (ln| x/4+1|-ln| x/4-1|)+x+C