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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3+x^2-x+1)/(x^3)

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

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

Solution steps

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

Expand

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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