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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-4x+3x^2)/(x^2)

\int \frac{1 - 4 x + 3 x ^{2}}{x ^{2}} d x

- \frac{1}{x} - 4 ln ∣ x ∣ + 3 x + C

Solution steps

\int \frac{1 - 4 x + 3 x ^{2}}{x ^{2}} d x

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

= \int \frac{1}{x ^{2}} - \frac{4 x}{x ^{2}} + \frac{3 x ^{2}}{x ^{2}} d x

Simplify

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

= \int \frac{1}{x ^{2}} - \frac{4}{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 \frac{1}{x ^{2}} d x - \int \frac{4}{x} d x + \int 3 d x

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int 3 d x = 3 x

= - \frac{1}{x} - 4 ln ∣ x ∣ + 3 x

Add a constant to the solution

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

f (x) = \frac{3}{x}, at x = 5

\frac{d}{d t} (\frac{1}{1 + t ^{12}})

f (x) = (2 x^{2} - 4) - 1

\int (x + sin (x)) d x

\frac{d ^{2}}{d x ^{2}} (e^{3 e^{x}})

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