completing the square

[icyhot2590]

integral of 2x/x^2+6x+13

(b/2)^2

2x +6 - 6/x^2+6x+13

integral of
2x +6 /x^2+6x+13 - 6/x^2+6x+113

i need help substituting the first part i got the second part

2x +6 /x^2+6x+13

this is what i got

6integral 1/x^2+6x+9-9+13 (6/2)^2=9

6integral1/(x+3)^2+4

a=2
u=x+3
du=dx

6/2arctan (x+3)/2 +c

 

[galactus]

What is it you're trying to integrate?. Your post is rather confounding.

I reckon it's \(\displaystyle \L\\\int\frac{2x}{x^{2}+6x+13}dx\)

If so, completing the square is a good way to go.

Completing the square on \(\displaystyle x^{2}+6x+13=(x+3)^{2}+4\)

Now, you could let \(\displaystyle u=x+3, \;\ du=dx, \;\ x=u-3\)

\(\displaystyle \L\\\int\frac{2(u-3)}{u^{2}+4}du=\int\frac{2u}{u^{2}+4}du-\int\frac{6}{u^{2}+4}du\)

Now integrate and resub.