What is the integral of (5-4x^3+2x^6)/(x^6) ?

The integral of (5-4x^3+2x^6)/(x^6) is -1/(x^5)+2/(x^2)+2x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5-4x^3+2x^6)/(x^6)

\int \frac{5 - 4 x ^{3} + 2 x ^{6}}{x ^{6}} d x

- \frac{1}{x ^{5}} + \frac{2}{x ^{2}} + 2 x + C

Solution steps

\int \frac{5 - 4 x ^{3} + 2 x ^{6}}{x ^{6}} d x

Expand

\frac{5 - 4 x ^{3} + 2 x ^{6}}{x ^{6}} : \frac{5}{x ^{6}} - \frac{4}{x ^{3}} + 2

= \int \frac{5}{x ^{6}} - \frac{4}{x ^{3}} + 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 \frac{5}{x ^{6}} d x - \int \frac{4}{x ^{3}} d x + \int 2 d x

\int \frac{5}{x ^{6}} d x = - \frac{1}{x ^{5}}

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

\int 2 d x = 2 x

= - \frac{1}{x ^{5}} - (- \frac{2}{x ^{2}}) + 2 x

Simplify

= - \frac{1}{x ^{5}} + \frac{2}{x ^{2}} + 2 x

Add a constant to the solution

= - \frac{1}{x ^{5}} + \frac{2}{x ^{2}} + 2 x + C

\int sin (2 x + 4) d x

\frac{d}{d x} (\frac{sin ( x )}{1 - 2 cos ( x )})

\frac{d y}{d x}, y = 7 x^{2} + 6 y x

\int e^{- θ} cos (8 θ) d θ

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

What is the integral of (5-4x^3+2x^6)/(x^6) ?
The integral of (5-4x^3+2x^6)/(x^6) is -1/(x^5)+2/(x^2)+2x+C