What is the integral of (3x^2+5x)/(x^3+x^2-x-1) ?

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

\int \frac{3 x ^{2} + 5 x}{x ^{3} + x ^{2} - x - 1} d x

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

Solution steps

\int \frac{3 x ^{2} + 5 x}{x ^{3} + x ^{2} - x - 1} d x

Take the partial fraction of

\frac{3 x ^{2} + 5 x}{x ^{3} + x ^{2} - x - 1} : \frac{1}{x + 1} + \frac{1}{( x + 1 ) ^{2}} + \frac{2}{x - 1}

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

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

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

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

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

Add a constant to the solution

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

\int 8 x + x d x

\int (2 x + 3)^{6} d x

\int (x^{\frac{1}{3}} - x + 5) d x

\int b a^{x} d x

\int (e^{x} - \frac{1}{7 x}) d x

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