What is the integral of (2x+3)/(x^2+4x+13) ?

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

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integral of (2x+3)/(x^2+4x+13)

\int \frac{2 x + 3}{x ^{2} + 4 x + 13} d x

ln x^{2} + 4 x + 13 - \frac{1}{3} arctan (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{2 x + 3}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int \frac{2 x + 3}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 9} d u

= \int \frac{2 u - 1}{u ^{2} + 9} d u

Expand

\frac{2 u - 1}{u ^{2} + 9} : \frac{2 u}{u ^{2} + 9} - \frac{1}{u ^{2} + 9}

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 u}{u ^{2} + 9} d u - \int \frac{1}{u ^{2} + 9} d u

\int \frac{2 u}{u ^{2} + 9} d u = ln u^{2} + 9

\int \frac{1}{u ^{2} + 9} d u = \frac{1}{3} arctan (\frac{u}{3})

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

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