What is the integral of (x^4+2x^3-7x^2+5x-2)/x ?

The integral of (x^4+2x^3-7x^2+5x-2)/x is (x^4)/4+(2x^3)/3-(7x^2)/2+5x-2ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^4+2x^3-7x^2+5x-2)/x

\int \frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} d x

\frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} d x

Expand

\frac{x ^{4} + 2 x ^{3} - 7 x ^{2} + 5 x - 2}{x} : x^{3} + 2 x^{2} - 7 x + 5 - \frac{2}{x}

= \int x^{3} + 2 x^{2} - 7 x + 5 - \frac{2}{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 x^{3} d x + \int 2 x^{2} d x - \int 7 x d x + \int 5 d x - \int \frac{2}{x} d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 5 d x = 5 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{7 x ^{2}}{2} + 5 x - 2 ln ∣ x ∣ + C

What is the integral of (x^4+2x^3-7x^2+5x-2)/x ?
The integral of (x^4+2x^3-7x^2+5x-2)/x is (x^4)/4+(2x^3)/3-(7x^2)/2+5x-2ln|x|+C