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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

\frac{1}{x + 1} - 2 ln ∣ x + 2 ∣ + 2 ln ∣ x + 1 ∣ - 3 (ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣ - \frac{1}{x + 1}) : \frac{4}{x + 1} - 5 ln ∣ x + 2 ∣ + 5 ln ∣ x + 1 ∣

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

Add a constant to the solution

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

\int x^{2021} d x

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

\int (x^{4} - 5 x)^{2} (4 x^{3} - 5) d x

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

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

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