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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{4}{3 ( x - 2 )} d x = \frac{4}{3} ln ∣ x - 2 ∣

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

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

= \frac{4}{3} ln ∣ x - 2 ∣ + \frac{2}{3} ln ∣ x + 1 ∣ - (- \frac{1}{x + 1})

Simplify

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

Add a constant to the solution

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

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