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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x^2+x^4)/(-4x^2)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

= (- \frac{1}{4}) \cdot \int \frac{2 x ^{2} + x ^{4}}{x ^{2}} d x

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

= - \frac{1}{4} (2 x + \frac{x ^{3}}{3}) + C

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