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

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

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

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

\frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3)) + C

Solution steps

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

Complete the square

x^{2} + 6 x + 18 : (x + 3)^{2} + 9

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

Apply u-substitution

: \int \frac{u - 3}{u ^{2} + 9} d u

= \int \frac{u - 3}{u ^{2} + 9} d u

Expand

\frac{u - 3}{u ^{2} + 9} : \frac{u}{u ^{2} + 9} - \frac{3}{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{u}{u ^{2} + 9} d u - \int \frac{3}{u ^{2} + 9} d u

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

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

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

Substitute back

u = x + 3

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

Simplify

\frac{1}{2} ln (x + 3)^{2} + 9 - arctan (\frac{x + 3}{3}) : \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3))

= \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 18 - arctan (\frac{1}{3} (x + 3)) + C

\frac{s + 1}{s ( s ^{2} + 4 s + 4 )}

\frac{d}{d x} (x csc (x))

x^{3} + 2 x

\int 9 sin (x) cos (x) d x

(2 e^{y} - x) y^{^{'}} = 1

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