calculus. integral (involves inverse trig), by completing the square perhaps?

Question

Evaluate the integral of (2x)/(x^2+6x+13). The answer given was: ln(x^2+6x+13) - 3arctan((x+3)/2) + C.

My initial thoughts were to complete the square for the denominator, but to integrate it with arctan would only work for a numerator of a number and the numerator in the problem has a variable, which I don't know how to encounter.. presently. I can't think of any other way of rewriting or splitting apart. Any nudges in the right direction, explanation of how the answer produces 2 terms, or working of the problem would be extremely helpful. Thanks!

Answer 1

Try writing
(2x)/(x^2+6x+13).
as
(2x+6)/(x^2+6x+13) - 6/(x^2+6x+13)

The integrals of each term are the same as each term in your answer. The first is just 1/u du where u = denominator, and in the second you can complete the square and use your arctan formula.

Answer 2

Forget about where the inverse circular functions come in and note that the derivative of the denominator can be made in the numerator.
i.e.
[(2x+6) - 6] / (x^2 + 6x +13)
the first part of the answer is now apparent and for the arctan part you need only to complete the square for the denominator and put it in a form such that the integral of the resulting fraction is involving the arctangent.

Answer 3

First, as you suggested, complete the square of the denominator.

x^2 + 6x + 13 = x^2 + 6x + 9 + 13 - 9
= (x - 3)^2 + 4

And then make the substitution

u = x -3
u - 3 = x
therefore, du = dx

Your new integral is now as follows, pulling out 2 as the constant.

2 * Integral ( (u-3)/[u^2 + 4])du

This can be split up into two integrals.

2 * [ Integral (u/[u^2 + 4])du - Integral (3/[u^2 + 4])du ]

The first integral can be solved by substitution (I'll just state the integral you sound like an expert enough to know how to solve it) and is equal to (1/2)ln|u^2 + 4|.
So we have:

2 * [ (1/2) ln|u^2 + 4| - Integral (3/[u^2 + 4]) du ]

Let's distribute that 2 on the outside, AND pull out that remaining constant in the remaining integral.

ln |u^2 + 4| - (2/3)*Integral (1/[u^2 + 4]) du

Now, we solve the integral using trig substitution.
Let u = 2tan(v)
du = 2sec^2(v) dv

Integral ( [1/[4tan^2(v) + 4]] * 2sec^2(v) dv )
(2/4) * Integral ( [1/(tan^2(v) + 1)] * sec^2(v) dv)
(1/2) * Integral [ (1/sec^2(v)) (sec^2(v) ] dv
(1/2) * Integral [1] dv
(1/2) * v + C

But u = 2tanv, so u/2 = tan v and v = arctan(u/2)

(1/2) arctan(u/2)

Therefore, plugging this back in for

ln |u^2 + 4| - (2/3)*Integral (1/[u^2 + 4]) du
We get
ln |u^2 + 4| - (2/3)[ (1/2) arctan(u/2) ] + C
ln |u^2 + 4| - (1/3) arctan(u/2) + C

But remember that we let u = x + 3, so our answer is

ln |(x+3)^2 + 4| - (1/3) arctan( (x+3)/2 ) + C

Answer 4

Make the substitution (x - 2) = 2*sin? dx = 2*cos? d? ? dx/?[4 - (x - 2)²] = ? 2*cos? d?/?(4 - 4*sin²?) = ? 2*cos? d?/(2cos?) = ? d? = ? + c = arcsin[(a million/2)(x - 2)] + c the limits are 2 to 3, so ==> arcsin[(a million/2)(a million)] - arcsin[(a million/2)(0)] = ?/6