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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= - \frac{1}{x ^{2}} - (- \frac{3}{x}) + 4 ln ∣ x ∣

Simplify

= - \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣ + C

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