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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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