What is the integral of (x^2+8)/(x^2) ?

The integral of (x^2+8)/(x^2) is x-8/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+8)/(x^2)

\int \frac{x ^{2} + 8}{x ^{2}} d x

x - \frac{8}{x} + C

Solution steps

\int \frac{x ^{2} + 8}{x ^{2}} d x

Expand

\frac{x ^{2} + 8}{x ^{2}} : 1 + \frac{8}{x ^{2}}

= \int 1 + \frac{8}{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 1 d x + \int \frac{8}{x ^{2}} d x

\int 1 d x = x

\int \frac{8}{x ^{2}} d x = - \frac{8}{x}

= x - \frac{8}{x}

Add a constant to the solution

= x - \frac{8}{x} + C

\frac{d}{d x} (x (x^{2} - 5)^{\frac{1}{3}})

\int y^{3} 1 + y^{4} d y

\int 29 tan^{6} (x) d x

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

\int - 7 cos (\frac{1}{7} x) d x

What is the integral of (x^2+8)/(x^2) ?
The integral of (x^2+8)/(x^2) is x-8/x+C