What is the integral of (-4/(x^3)-8/(x^5)) ?

The integral of (-4/(x^3)-8/(x^5)) is 2/(x^2)+2/(x^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-4/(x^3)-8/(x^5))

\int (- \frac{4}{x ^{3}} - \frac{8}{x ^{5}}) d x

\frac{2}{x ^{2}} + \frac{2}{x ^{4}} + C

Solution steps

\int - \frac{4}{x ^{3}} - \frac{8}{x ^{5}} 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{4}{x ^{3}} d x - \int \frac{8}{x ^{5}} d x

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

\int \frac{8}{x ^{5}} d x = - \frac{2}{x ^{4}}

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

Simplify

= \frac{2}{x ^{2}} + \frac{2}{x ^{4}}

Add a constant to the solution

= \frac{2}{x ^{2}} + \frac{2}{x ^{4}} + C

\int \frac{x}{x + x ^{\frac{4}{5}}} d x

\int_{3}^{6} 8 d x

\int - 5 π sin (π x) d x

\int \frac{1}{( 1 + x ) ^{3}} d x

- \frac{2}{x ^{3}}

What is the integral of (-4/(x^3)-8/(x^5)) ?
The integral of (-4/(x^3)-8/(x^5)) is 2/(x^2)+2/(x^4)+C