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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{x} + x + C

Solution steps

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

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

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

\int 1 d x = x

= - \frac{1}{x} + x

Add a constant to the solution

= - \frac{1}{x} + x + C

x e^{y} d y + \frac{x ^{2} + 1}{y} d x = 0

\frac{d}{d x} (\frac{1 - 2 x}{1 + 2 x})

\frac{d}{d x} (x^{2} sin (x) - x cos (x))

(3 x^{3} + 2 x) (x - 2) (x + 4)

cos (x) d x - sin (y) d y = 0

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