CALC 2: trig sub help, please? integrate: (x+4)/(X^2+2X+4)?

Question

Hi guys. Looking at the integral, I initially thought using partial fractions would make it a piece of cake, but obviously x^2+2x+4 cannot be factored. So, I completed the square, giving us int of (x+4)/[(x+1)^2+3]. This looks as if we should use trig sub... could someone please show me their method to integrate this function? I'm 99% sure I solved it COMPETELY wrong. Please help! Thank you =D

WOW thank you both! =D
radne0, the fact that you've saved me twice today officially makes you my HERO :)

Answer 1

∫(x+4)/(x^2+2x+4)
You're correct it's not factorable..

∫(x+4)/(x^2+2x+4) dx

Well if we let u = x^2+2x+4... du/dx will be 2x+2 Let's manipulate this so we have just that in the numerator and split anything extra into another integral :)

Multiply by 2/2 but leave the denominator 2 outside integral

1/2∫2(x+4)/(x^2+2x+4) dx
1/2∫(2x+8)/(x^2+2x+4) dx

Now again 2x+2 is du/dx.. so let's write 2x+8 as 2x+2 +6
1/2∫(2x+2+6)/(x^2+2x+4)

And split it into two integrals :)
1/2∫(2x+2+6)/(x^2+2x+4) = 1/2[ ∫(2x+2)/(x^2+2x+4) + ∫6/(x^2+2x+4)dx]

The first is easy enough, just u substitution with du/u and the second will be trig substitution as you suggested.

First one, ∫(2x+2)/(x^2+2x+4)
u = x^2+2x+4
du/dx = 2x +2
dx = du/(2x+2)
∫du/u
ln |u| + c
ln |x^2+2x+4|...

Second one, ∫6/(x^2+2x+4)dx

start off by completing the square as you did

∫6/[(x+1)^2+3]
6∫1/[(x+1)^2+3]

At this point if you're allowed there's probably a formula for
du/(u^2+a^2) in your tables.. but if not.

Let x +1 = √3 tant(t) and therefore
dx/dt = √3 sec^2(t)
dx = √3 sec^2(t) dt

Integral becomes
6 ∫√3 sec^2(t)/[3 tan^2+3]

factor out 3 and remember tan^2(t) = sec^2(t)
(6√3/3) ∫sec^2(t)/[ tan^2(t)+1]
(2√3)∫sec^2(t)/[ sec^2(t)]

Simplifies to just
2 √3/3∫1
2 √3/3 * t +c

We need to get this in terms of x.. so go back to our original substitution and solve for tan(t) and take arctan :)
x +1 = √3 tant(t)
(x+1)/√3 = tan(t)
t = arctan((x+1)/√3)
[2√3]/3 arctan((x+1)/√3) + C

put the two together
(1/2) [ln |x^2+2x+4| + (2√3)/3 arctan((x+1)/√3)] + C
(1/2) ln |x^2+2x+4| + √3 arctan((x+1)/√3) + C

Oh, because x^2+2x+4 is always > 0 you don't need the absolute value
(1/2) ln (x^2+2x+4) + √3 arctan((x+1)/√3) + C

Answer 2

Hi

integral (x+4)/(x^2+2x+4) dx

when I first saw this integral I looked the derivative of the bottom, which is 2x+2, or 2(x+1). So what I did is this:

integral (x+1+3)/(x^2+2x+4) dx
= integral (x+1)/(x^2+2x+4) + integral 3/(x^2+2x+4) dx
For the first part, let u = x^2+2x+4, so that du = 2x+2 dx --> du/2 = x+1 dx.
The first integral becomes 1/2 integral 1/u du, which equals 1/2 ln(x^2+2x+4).

The second integral can be done with complete the square like you said.

integral 3/[(x+1)^2 +3] dx
Let w = x+1 and dw = dx
integral 3/(w^2+3) dw
divide top and bottom by 3
integral 1/[(w/sqrt(3))^2 + 1] dw
This is the integral of arctangent, which evaulates into sqrt(3) arctan(w/sqrt(3)) = sqrt(3)arctan[(x+1)/sqrt(3)]

So putting it all together, you get 1/2 ln(x^2+2x+4) + sqrt(3)arctan[(x+1)/sqrt(3)] + C.

Answer 3

Not as bad as you make it out to be.
Let U= the quadratic and dU= (2x+2)dx. With some shuffling around, (x+4)= 2(x+4)/2= (2x+2+6)/2 . So (x+4)dx = dU/2 + 3dx
Your problem is now INT [1/2 dU/U] + INT [(3dx/(quadratic) ]. The first term is a ln(U) term, while the second is a standard form.