What is the integral of (3x^2+9x^4-x)/(x^2) ?

The integral of (3x^2+9x^4-x)/(x^2) is 3x+3x^3-ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x^2+9x^4-x)/(x^2)

\int \frac{3 x ^{2} + 9 x ^{4} - x}{x ^{2}} d x

3 x + 3 x^{3} - ln ∣ x ∣ + C

Solution steps

\int \frac{3 x ^{2} + 9 x ^{4} - x}{x ^{2}} d x

Expand

\frac{3 x ^{2} + 9 x ^{4} - x}{x ^{2}} : 3 + 9 x^{2} - \frac{1}{x}

= \int 3 + 9 x^{2} - \frac{1}{x} 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 3 d x + \int 9 x^{2} d x - \int \frac{1}{x} d x

\int 3 d x = 3 x

\int 9 x^{2} d x = 3 x^{3}

\int \frac{1}{x} d x = ln ∣ x ∣

= 3 x + 3 x^{3} - ln ∣ x ∣

Add a constant to the solution

= 3 x + 3 x^{3} - ln ∣ x ∣ + C

What is the integral of (3x^2+9x^4-x)/(x^2) ?
The integral of (3x^2+9x^4-x)/(x^2) is 3x+3x^3-ln|x|+C