What is the integral of (x^2-1)/(2x^2-32) ?

The integral of (x^2-1)/(2x^2-32) is 1/2 x-15/16 (ln| x/4+1|-ln| x/4-1|)+C

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integral of (x^2-1)/(2x^2-32)

\int \frac{x ^{2} - 1}{2 x ^{2} - 32} d x

\frac{1}{2} x - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

Solution steps

\int \frac{x ^{2} - 1}{2 x ^{2} - 32} d x

Long division

\frac{x ^{2} - 1}{2 x ^{2} - 32} : \frac{1}{2} + \frac{15}{2 x ^{2} - 32}

= \int \frac{1}{2} + \frac{15}{2 x ^{2} - 32} 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{1}{2} d x + \int \frac{15}{2 x ^{2} - 32} d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{15}{2 x ^{2} - 32} d x = - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

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

Add a constant to the solution

= \frac{1}{2} x - \frac{15}{16} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

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What is the integral of (x^2-1)/(2x^2-32) ?
The integral of (x^2-1)/(2x^2-32) is 1/2 x-15/16 (ln| x/4+1|-ln| x/4-1|)+C