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

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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

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