What is the integral of (7x+2)/(x^2-4) ?

The integral of (7x+2)/(x^2-4) is 7/2 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 (7x+2)/(x^2-4)

\int \frac{7 x + 2}{x ^{2} - 4} d x

\frac{7}{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{7 x + 2}{x ^{2} - 4} d x

Expand

\frac{7 x + 2}{x ^{2} - 4} : \frac{7 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{7 x}{x ^{2} - 4} d x + \int \frac{2}{x ^{2} - 4} d x

\int \frac{7 x}{x ^{2} - 4} d x = \frac{7}{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{7}{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{7}{2} ln x^{2} - 4 - \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 + C

\frac{d}{d x} (\frac{( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} )})

\int e^{x} sin^{2} (x) d x

\frac{d y}{d x} = \frac{x}{y}

\frac{10 x + 4 x ^{2}}{( 5 + 4 x ) ^{2}}

\frac{d}{d x} (3 3 tan (x^{2}))

What is the integral of (7x+2)/(x^2-4) ?
The integral of (7x+2)/(x^2-4) is 7/2 ln|x^2-4|-1/2 ln| x/2+1|+1/2 ln| x/2-1|+C