What is the integral of (9x^2-6)/(x^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9x^2-6)/(x^2)

\int \frac{9 x ^{2} - 6}{x ^{2}} d x

9 x + \frac{6}{x} + C

Solution steps

\int \frac{9 x ^{2} - 6}{x ^{2}} d x

Expand

\frac{9 x ^{2} - 6}{x ^{2}} : 9 - \frac{6}{x ^{2}}

= \int 9 - \frac{6}{x ^{2}} 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 9 d x - \int \frac{6}{x ^{2}} d x

\int 9 d x = 9 x

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

= 9 x - (- \frac{6}{x})

Simplify

= 9 x + \frac{6}{x}

Add a constant to the solution

= 9 x + \frac{6}{x} + C

\frac{\partial}{\partial x} (\frac{- 8 x + 5}{2 y ^{2} + 1})

y^{^{'}} + \frac{1}{x} y = 7 x^{2}

\frac{d}{d x} (ln (x) - \frac{1}{x})

x = 0 \sum \infty sin (x)

\int x x - 2 d x

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