Question

integrate 1/(x^2+6x+18)

Answer 1

partial fraction decomp... see attached

Answer 2

oops, may be something different. just a sec.

Answer 3

i think it is partial fractions but I have no idea what to do with the denomenator

Answer 4

it's an irreducible quadratic

Answer 5

i think you have to complete the square... hold on

Answer 6

try this

Answer 7

\[x^2+6x+18 = (x+3)^2 + 9\] let u = x+3 then you can use an inverse trig fucntion to integrate

\[\int\limits \frac{ 1 }{ x^2+6x+18 }\,dx=\int\limits \frac{ 1 }{ \left( x+3 \right)^2+9}\,dx=\int\limits \frac{ 1 }{ u^2+9}\,du\]

Answer 8

then can i just use the formula \[\int\limits dx/(x^2+a^2) = 1/a \arctan (x/a) + C\]

Answer 9

wrong trig function...
have a look

Answer 10

my bad... you got it.

Answer 11

be careful how you write in here, though. better to use \frac then just \

1/n arctan(n/a) or \(\large{\frac{1}{n} \arctan \left(\frac{n}{a}\right)}\)

Answer 12

i thought you had the arctan on the bottom

Answer 13

you good to go?

Answer 14

yes, thanks a lot

Answer 15

you're welcome!