What is the integral of (x^3-3x+2)/((x+1)(x-1)^3) ?

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

\int \frac{x ^{3} - 3 x + 2}{( x + 1 ) ( x - 1 ) ^{3}} d x

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

Solution steps

\int \frac{x ^{3} - 3 x + 2}{( x + 1 ) ( x - 1 ) ^{3}} d x

Simplify

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

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

Expand

\frac{x + 2}{( x + 1 ) ( x - 1 )} : \frac{x}{( x + 1 ) ( x - 1 )} + \frac{2}{( x + 1 ) ( x - 1 )}

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

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

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

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

Add a constant to the solution

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

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

\int_{0}^{\frac{π}{2}} sin (2 x - π) d x

k = 1 \sum \infty \frac{1}{k !}

y = \frac{5}{( 2 x ^{3} )} + 2 cos (x)

\int \frac{ln ( x ^{9} )}{x} d x

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