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

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

Popular >

integral of (4x+3)/(x^3+2x^2+x)

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int \frac{1}{x e ^{x}} d x

\int (2 + 3 x^{2}) cos (3 x) d x

\int \frac{( 3 x - 2 )}{x ^{2} + 2 x + 17} d x

\int \frac{1}{x + 2 + x + 2} d x

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

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