What is the integral of (x+18)/(x^2+6x+13) ?

The integral of (x+18)/(x^2+6x+13) is 1/2 ln|x^2+6x+13|+15/2 arctan((x+3)/2)+C

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integral of (x+18)/(x^2+6x+13)

\int \frac{x + 18}{x ^{2} + 6 x + 13} d x

\frac{1}{2} ln x^{2} + 6 x + 13 + \frac{15}{2} arctan (\frac{x + 3}{2}) + C

Solution steps

\int \frac{x + 18}{x ^{2} + 6 x + 13} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 30 u + 229} d u

= \int \frac{u}{u ^{2} - 30 u + 229} d u

Complete the square

u^{2} - 30 u + 229 : (u - 15)^{2} + 4

= \int \frac{u}{( u - 15 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 15}{v ^{2} + 4} d v

= \int \frac{v + 15}{v ^{2} + 4} d v

Expand

\frac{v + 15}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{15}{v ^{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{v}{v ^{2} + 4} d v + \int \frac{15}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

\int \frac{15}{v ^{2} + 4} d v = \frac{15}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{15}{2} arctan (\frac{v}{2})

Substitute back

= \frac{1}{2} ln (x + 18 - 15)^{2} + 4 + \frac{15}{2} arctan (\frac{x + 18 - 15}{2})

Simplify

\frac{1}{2} ln (x + 18 - 15)^{2} + 4 + \frac{15}{2} arctan (\frac{x + 18 - 15}{2}) : \frac{1}{2} ln x^{2} + 6 x + 13 + \frac{15}{2} arctan (\frac{x + 3}{2})

= \frac{1}{2} ln x^{2} + 6 x + 13 + \frac{15}{2} arctan (\frac{x + 3}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 13 + \frac{15}{2} arctan (\frac{x + 3}{2}) + C

n = 1 \sum \infty \frac{1}{( n + 2 )}

\int x (2 - x) d x

\frac{\partial}{\partial x} (\frac{18 x}{( y + 9 z ) ^{3}})

\int_{0}^{\infty} x^{0} e^{- x} d x

\frac{\partial}{\partial x} (x y (4 - y))

What is the integral of (x+18)/(x^2+6x+13) ?
The integral of (x+18)/(x^2+6x+13) is 1/2 ln|x^2+6x+13|+15/2 arctan((x+3)/2)+C