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

The integral of (x+5)/((x+4)(x^2-1)) is 1/15 ln|x+4|-2/3 ln|x+1|+3/5 ln|x-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+5)/((x+4)(x^2-1))

\int \frac{x + 5}{( x + 4 ) ( x ^{2} - 1 )} d x

\frac{1}{15} ln ∣ x + 4 ∣ - \frac{2}{3} ln ∣ x + 1 ∣ + \frac{3}{5} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x + 5}{( x + 4 ) ( x ^{2} - 1 )} d x

Take the partial fraction of

\frac{x + 5}{( x + 4 ) ( x ^{2} - 1 )} : \frac{1}{15 ( x + 4 )} - \frac{2}{3 ( x + 1 )} + \frac{3}{5 ( x - 1 )}

= \int \frac{1}{15 ( x + 4 )} - \frac{2}{3 ( x + 1 )} + \frac{3}{5 ( x - 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{1}{15 ( x + 4 )} d x - \int \frac{2}{3 ( x + 1 )} d x + \int \frac{3}{5 ( x - 1 )} d x

\int \frac{1}{15 ( x + 4 )} d x = \frac{1}{15} ln ∣ x + 4 ∣

\int \frac{2}{3 ( x + 1 )} d x = \frac{2}{3} ln ∣ x + 1 ∣

\int \frac{3}{5 ( x - 1 )} d x = \frac{3}{5} ln ∣ x - 1 ∣

= \frac{1}{15} ln ∣ x + 4 ∣ - \frac{2}{3} ln ∣ x + 1 ∣ + \frac{3}{5} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{15} ln ∣ x + 4 ∣ - \frac{2}{3} ln ∣ x + 1 ∣ + \frac{3}{5} ln ∣ x - 1 ∣ + C

\int 25 - x^{2} (2 x) d x

\frac{\partial}{\partial x} (\frac{- t ^{2}}{( 1 - x ^{2} )})

y^{^{'}} - 2 y = x + 5

x \to \frac{5 π}{2} - lim (sec (x))

\int \frac{y}{z ^{2}} d z

What is the integral of (x+5)/((x+4)(x^2-1)) ?
The integral of (x+5)/((x+4)(x^2-1)) is 1/15 ln|x+4|-2/3 ln|x+1|+3/5 ln|x-1|+C