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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = e^{- 2 y} tan^{2} (x)

x^{^{''}} + x^{^{'}} - 2 x = 0

y^{^{'}} = - \frac{y}{x}

x^{^{'''}} + x^{^{''}} + 3 x^{^{'}} - 5 x = 0

e^{x^{2} + 1}

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