What is the integral of (3x^2+12x-9)/(x^4) ?

The integral of (3x^2+12x-9)/(x^4) is -3/x-6/(x^2)+3/(x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2+12x-9)/(x^4)

\int \frac{3 x ^{2} + 12 x - 9}{x ^{4}} d x

- \frac{3}{x} - \frac{6}{x ^{2}} + \frac{3}{x ^{3}} + C

Solution steps

\int \frac{3 x ^{2} + 12 x - 9}{x ^{4}} d x

Expand

\frac{3 x ^{2} + 12 x - 9}{x ^{4}} : \frac{3}{x ^{2}} + \frac{12}{x ^{3}} - \frac{9}{x ^{4}}

= \int \frac{3}{x ^{2}} + \frac{12}{x ^{3}} - \frac{9}{x ^{4}} 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{3}{x ^{2}} d x + \int \frac{12}{x ^{3}} d x - \int \frac{9}{x ^{4}} d x

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

\int \frac{12}{x ^{3}} d x = - \frac{6}{x ^{2}}

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

= - \frac{3}{x} - \frac{6}{x ^{2}} - (- \frac{3}{x ^{3}})

Simplify

= - \frac{3}{x} - \frac{6}{x ^{2}} + \frac{3}{x ^{3}}

Add a constant to the solution

= - \frac{3}{x} - \frac{6}{x ^{2}} + \frac{3}{x ^{3}} + C

\int cos (\frac{π x}{12}) d x

\int \frac{s ^{2} + 2 s}{s ^{3} + 3 s ^{2} + 1} d s

9 sin (x)

\int \frac{cos ( 5 x )}{1 - sin ( 5 x )} d x

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

What is the integral of (3x^2+12x-9)/(x^4) ?
The integral of (3x^2+12x-9)/(x^4) is -3/x-6/(x^2)+3/(x^3)+C