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

The integral of (x^4+x^3-2x^2+2x+8)/(x^2+4) is (x^3)/3+1/2 x^2-6x+16arctan(x/2)-ln|x^2+4|+2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4+x^3-2x^2+2x+8)/(x^2+4)

\int \frac{x ^{4} + x ^{3} - 2 x ^{2} + 2 x + 8}{x ^{2} + 4} d x

\frac{x ^{3}}{3} + \frac{1}{2} x^{2} - 6 x + 16 arctan (\frac{x}{2}) - ln x^{2} + 4 + 2 + C

Solution steps

\int \frac{x ^{4} + x ^{3} - 2 x ^{2} + 2 x + 8}{x ^{2} + 4} d x

Expand

\frac{x ^{4} + x ^{3} - 2 x ^{2} + 2 x + 8}{x ^{2} + 4} : \frac{x ^{4}}{x ^{2} + 4} + \frac{x ^{3}}{x ^{2} + 4} - \frac{2 x ^{2}}{x ^{2} + 4} + \frac{2 x}{x ^{2} + 4} + \frac{8}{x ^{2} + 4}

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{x ^{4}}{x ^{2} + 4} d x + \int \frac{x ^{3}}{x ^{2} + 4} d x - \int \frac{2 x ^{2}}{x ^{2} + 4} d x + \int \frac{2 x}{x ^{2} + 4} d x + \int \frac{8}{x ^{2} + 4} d x

\int \frac{x ^{4}}{x ^{2} + 4} d x = 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2}))

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

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

\int \frac{2 x}{x ^{2} + 4} d x = ln x^{2} + 4

\int \frac{8}{x ^{2} + 4} d x = 4 arctan (\frac{x}{2})

= 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) + \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4) - 4 (- arctan (\frac{x}{2}) + \frac{x}{2}) + ln x^{2} + 4 + 4 arctan (\frac{x}{2})

Simplify

8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) + \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4) - 4 (- arctan (\frac{x}{2}) + \frac{x}{2}) + ln x^{2} + 4 + 4 arctan (\frac{x}{2}) : \frac{x ^{3}}{3} + \frac{1}{2} x^{2} - 6 x + 16 arctan (\frac{x}{2}) - ln x^{2} + 4 + 2

= \frac{x ^{3}}{3} + \frac{1}{2} x^{2} - 6 x + 16 arctan (\frac{x}{2}) - ln x^{2} + 4 + 2

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{1}{2} x^{2} - 6 x + 16 arctan (\frac{x}{2}) - ln x^{2} + 4 + 2 + C

\int \frac{1}{- 2 + 3 x - 5 x ^{2}} d x

\int \frac{3 x ^{8} + 1}{x ^{9} + 3 x} d x

\int x^{- 3} e^{x} d x

\int \frac{x - 3}{2 x + 1} d x

\int \frac{e ^{5 x}}{x} d x

What is the integral of (x^4+x^3-2x^2+2x+8)/(x^2+4) ?
The integral of (x^4+x^3-2x^2+2x+8)/(x^2+4) is (x^3)/3+1/2 x^2-6x+16arctan(x/2)-ln|x^2+4|+2+C