What is the integral of (x^2-4x+3)/(x^3+2x^2+x) ?

The integral of (x^2-4x+3)/(x^3+2x^2+x) is 3ln|x|-2ln|x+1|+8/(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^2-4x+3)/(x^3+2x^2+x)

\int \frac{x ^{2} - 4 x + 3}{x ^{3} + 2 x ^{2} + x} d x

3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1} + C

Solution steps

\int \frac{x ^{2} - 4 x + 3}{x ^{3} + 2 x ^{2} + x} d x

Take the partial fraction of

\frac{x ^{2} - 4 x + 3}{x ^{3} + 2 x ^{2} + x} : \frac{3}{x} - \frac{2}{x + 1} - \frac{8}{( x + 1 ) ^{2}}

= \int \frac{3}{x} - \frac{2}{x + 1} - \frac{8}{( 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{3}{x} d x - \int \frac{2}{x + 1} d x - \int \frac{8}{( x + 1 ) ^{2}} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{8}{( x + 1 ) ^{2}} d x = - \frac{8}{x + 1}

= 3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ - (- \frac{8}{x + 1})

Simplify

= 3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1}

Add a constant to the solution

= 3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + \frac{8}{x + 1} + C

\frac{\partial}{\partial x} (x + y + x y)

z \to 2 lim (7 z^{2} + 14 z - 7)

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

y = (3 x^{2} + 2) (4 x - 3)

\frac{5}{s - 3} - \frac{4}{s + 6}

What is the integral of (x^2-4x+3)/(x^3+2x^2+x) ?
The integral of (x^2-4x+3)/(x^3+2x^2+x) is 3ln|x|-2ln|x+1|+8/(x+1)+C