What is the integral of (2x^4-18)/(x^2+3) ?

The integral of (2x^4-18)/(x^2+3) is (2x^3)/3-6x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^4-18)/(x^2+3)

\int \frac{2 x ^{4} - 18}{x ^{2} + 3} d x

\frac{2 x ^{3}}{3} - 6 x + C

Solution steps

\int \frac{2 x ^{4} - 18}{x ^{2} + 3} d x

Expand

\frac{2 x ^{4} - 18}{x ^{2} + 3} : \frac{2 x ^{4}}{x ^{2} + 3} - \frac{18}{x ^{2} + 3}

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{2 x ^{4}}{x ^{2} + 3} d x - \int \frac{18}{x ^{2} + 3} d x

\int \frac{2 x ^{4}}{x ^{2} + 3} d x = \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3})

\int \frac{18}{x ^{2} + 3} d x = 63 arctan (\frac{x}{3})

= \frac{2 x ^{3}}{3} - 6 x + 63 arctan (\frac{x}{3}) - 63 arctan (\frac{x}{3})

Simplify

= \frac{2 x ^{3}}{3} - 6 x

Add a constant to the solution

= \frac{2 x ^{3}}{3} - 6 x + C

\int e^{- 0.15 x} d x

\int \frac{1 + y}{1 - y} d y

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

\int (5 x^{2} - 4 x + 3) d x

\int 19 t e^{t} d t

What is the integral of (2x^4-18)/(x^2+3) ?
The integral of (2x^4-18)/(x^2+3) is (2x^3)/3-6x+C