What is the integral of (1-2x^2-4/(x^{-3)})x^{-3} ?

The integral of (1-2x^2-4/(x^{-3)})x^{-3} is -1/(2x^2)-2ln|x|-4x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-2x^2-4/(x^{-3)})x^{-3}

\int (1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} d x

- \frac{1}{2 x ^{2}} - 2 ln ∣ x ∣ - 4 x + C

Solution steps

\int (1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} d x

Simplify

(1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} : \frac{1 - 2 x ^{2} - 4 x ^{3}}{x ^{3}}

= \int \frac{1 - 2 x ^{2} - 4 x ^{3}}{x ^{3}} d x

Expand

\frac{1 - 2 x ^{2} - 4 x ^{3}}{x ^{3}} : \frac{1}{x ^{3}} - \frac{2}{x} - 4

= \int \frac{1}{x ^{3}} - \frac{2}{x} - 4 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{1}{x ^{3}} d x - \int \frac{2}{x} d x - \int 4 d x

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 4 d x = 4 x

= - \frac{1}{2 x ^{2}} - 2 ln ∣ x ∣ - 4 x

Add a constant to the solution

= - \frac{1}{2 x ^{2}} - 2 ln ∣ x ∣ - 4 x + C

\int (7 x^{2} - 12 x) e^{2 x} d x

x \to - 2 - lim (- 3 x^{2} + 2)

y^{^{'}} = \frac{1}{2} + \frac{1}{4} sin (t) - \frac{y}{50}

\int \frac{a}{b ^{3 θ}} d θ

What is the integral of (1-2x^2-4/(x^{-3)})x^{-3} ?
The integral of (1-2x^2-4/(x^{-3)})x^{-3} is -1/(2x^2)-2ln|x|-4x+C