What is the integral of (x-1)/(x^2-4) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-1)/(x^2-4)

\int \frac{x - 1}{x ^{2} - 4} d x

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

Solution steps

\int \frac{x - 1}{x ^{2} - 4} d x

Expand

\frac{x - 1}{x ^{2} - 4} : \frac{x}{x ^{2} - 4} - \frac{1}{x ^{2} - 4}

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} - 4} d x - \int \frac{1}{x ^{2} - 4} d x

\int \frac{x}{x ^{2} - 4} d x = \frac{1}{2} ln x^{2} - 4

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{9}{x ^{2} - 16} d x

x \to 2 - lim (x^{3})

\int \frac{3}{5 + 3 x} d x

y^{^{''}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f (x) = 4 x

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