What is the integral of ((5x-2)/(x^2-4)) ?

The integral of ((5x-2)/(x^2-4)) is 5/2 ln|x^2-4|+1/2 ln| x/2+1|-1/2 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 ((5x-2)/(x^2-4))

\int (\frac{5 x - 2}{x ^{2} - 4}) d x

\frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

Solution steps

\int \frac{5 x - 2}{x ^{2} - 4} d x

Expand

\frac{5 x - 2}{x ^{2} - 4} : \frac{5 x}{x ^{2} - 4} - \frac{2}{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{5 x}{x ^{2} - 4} d x - \int \frac{2}{x ^{2} - 4} d x

\int \frac{5 x}{x ^{2} - 4} d x = \frac{5}{2} ln x^{2} - 4

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

= \frac{5}{2} ln x^{2} - 4 - (- \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1)

Simplify

= \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1

Add a constant to the solution

= \frac{5}{2} ln x^{2} - 4 + \frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 + C

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

\int (7 + tan^{2} (θ)) d θ

\int \frac{x ^{3} + 2 x ^{2} - 7 x}{x} d x

\int x (x + 1) d x

\int \frac{2 x + 1}{x ^{2} - 5 x + 6} d x

What is the integral of ((5x-2)/(x^2-4)) ?
The integral of ((5x-2)/(x^2-4)) is 5/2 ln|x^2-4|+1/2 ln| x/2+1|-1/2 ln| x/2-1|+C