For the given integral: `int 2x/(x^2+6x+13) dx` , we may apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .
`int 2x/(x^2+6x+13) dx =2 int x/(x^2+6x+13) dx`
To be able to evaluate this, we apply completing the square on `x^2+6x+13` .
The `x^2+6x+13` resembles `ax^2+bx+c` where:
`a= 1` and `b =6` that we can plug-into `(-b/(2a))^2` .
`(-b/(2a))^2= (-(6)/(2*1))^2`
`= (-6/2)^2`
`= (-3)^2`
`=9`
To complete the square, we add and subtract 9:
`x^2+6x+13 +9 -9`
Group them as: `(x^2+6x+9)-9+13`
Simplify: `(x^2+6x+9)+4`
Apply factoring for the perfect square trinomial: `x^2+6x+9 = (x+3)^2`
`(x^2+6x+9)+4=(x+3)^2 + 4`
Which means `x^2+6x+13 =(x+3)^2 + 4` then the integral becomes:
`2 int x/sqrt(x^2+6x+13) dx =2 int x/((x+3)^2 + 4) dx`
For the integral part, we apply u-substitution by letting:
`u = x+3` then `x= u-3` and `du =dx`
Then,
`2 int x/((x+3)^2 + 4) dx= 2 int (u-3)/(u^2 + 4) du`
Apply the basic integration property: : `int (u+v) dx = int (u) dx + int (v) dx` .
`2 int (u-3)/(u^2 + 4) du=2 [int u/(u^2 + 4) du - int 3/(u^2 + 4) du]`
For the integration of `int u/(u^2 + 4) du` , let:
`v=u^2+4` then `dv =2u du` or `(dv)/2 = u du` .
Then,
`int u/(u^2 + 4) du = int ((dv)/2)/(v)`
`= 1/2 int (dv)/(v)`
` = 1/2ln|v|+C`
Plug-in `v= u^2+4,` we get: `int u/(u^2 + 4) du =1/2ln|u^2+4|+C`
For the second integration: `- int 3/(u^2 + 4) du` , we follow the basic integration formula for inverse tangent function:
`int (du)/(u^2+a^2) = 1/a arctan(u/a)+C`
Then,
`- int 3/(u^2 + 4) du =-3 int (du)/(u^2 + 2^2)`
`= -3 *1/2arctan(u/2)+C`
`=-3/2 arctan(u/2)+C`
Combine the results, we get:
`2 [int (u/(u^2 + 4) du - int 3/(u^2 + 4) du]`
`=2*[ 1/2ln|u^2+4|-3/2arctan(u/2)]+C`
`= ln|u^2+4| - 3arctan(u/2)+C`
Plug-in `u=x+3` to solve for the final answer:
`int 2x/(x^2+6x+13) dx= ln|(x+3)^2+4| - 3arctan((x+3)/2)+C`
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