What is the integral of (x-2)/(x^2+4x+13) ?

The integral of (x-2)/(x^2+4x+13) is 1/2 ln| 1/9 (x+2)^2+1|-4/3 arctan(1/3 (x+2))+C

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integral of (x-2)/(x^2+4x+13)

\int \frac{x - 2}{x ^{2} + 4 x + 13} d x

\frac{1}{2} ln \frac{1}{9} (x + 2)^{2} + 1 - \frac{4}{3} arctan (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{x - 2}{x ^{2} + 4 x + 13} d x

Expand

\frac{x - 2}{x ^{2} + 4 x + 13} : \frac{x}{x ^{2} + 4 x + 13} - \frac{2}{x ^{2} + 4 x + 13}

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} + 4 x + 13} d x - \int \frac{2}{x ^{2} + 4 x + 13} d x

\int \frac{x}{x ^{2} + 4 x + 13} d x = \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2))

\int \frac{2}{x ^{2} + 4 x + 13} d x = \frac{2}{3} arctan (\frac{1}{3} (x + 2))

= \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) - \frac{2}{3} arctan (\frac{1}{3} (x + 2))

Simplify

\frac{1}{2} ln \frac{( x + 2 ) ^{2}}{9} + 1 - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) - \frac{2}{3} arctan (\frac{1}{3} (x + 2)) : \frac{1}{2} ln \frac{1}{9} (x + 2)^{2} + 1 - \frac{4}{3} arctan (\frac{1}{3} (x + 2))

= \frac{1}{2} ln \frac{1}{9} (x + 2)^{2} + 1 - \frac{4}{3} arctan (\frac{1}{3} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln \frac{1}{9} (x + 2)^{2} + 1 - \frac{4}{3} arctan (\frac{1}{3} (x + 2)) + C

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What is the integral of (x-2)/(x^2+4x+13) ?
The integral of (x-2)/(x^2+4x+13) is 1/2 ln| 1/9 (x+2)^2+1|-4/3 arctan(1/3 (x+2))+C