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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\int 2 x + 3 x^{1.7} d x

\int - 6 + 10 x^{3} - 8 x^{5} + 6 x^{7} d x

\frac{1}{1 - x ^{2}}

\int \frac{x + 1}{3 - x} d x

\int y (b - a y^{2}) d y

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