Your denominator might look scary, but it's actually a quadratic function in disguise, and we love those because they're incredibly simple to work.
If you say
u = ex
u2 = (ex)2 = e2x
So when you look at your denominator:
e2x + 5ex + 4
what you really have is
u2 + 5u + 4
which you can recognize immediately as
Your numerator is a little more simple
-17ex - 20
-17u - 20
So your integrand winds up looking like this:
(-17u - 20)/((u+4)(u+1))
Before moving forward, let's break this up using partial fraction decomposition (go ahead and facepalm, we all know students hate this).
(-17u - 20)/((u+4)(u+1)) = A/(u+4) + B/(u+1)
A(u+1) + B(u+4) = -17u - 20
Au + A + Bu + 4B = -17u - 20
(A+B)u + (A + 4B) = -17u - 20
A+B = -17
A+4B = -20
-A - B = 17
A + 4B = -20
3B = -3
B = -1
A = -16
So, your integrand looks like:
-16/(u+4) - 1/(u+1)
If we go back to the beginning, we said that u = ex
so let's substitute that back in.
-16/(ex + 4) - 1/(ex + 1)
∫-16/(ex + 4) - 1/(ex + 1) dx
Factor out ex from each denominator
∫-16/ex(1 + 4/ex) - 1/ex(1 + 1/ex) dx
Do a u-substitution for each term.
u1 = 1 + 4/ex
du1 = -4/ex dx
-du1 = 4/ex dx
u2 = 1 + 1/ex
du2 = -1/ex dx
-du2 = 1/ex dx
∫4/u1 du1 + ∫1/u2 du2
= 4*ln(u1) + ln(u2) + C
= 4*ln(1 + 4/ex) + ln(1 + 1/ex) + C
= 4*ln(e-x(ex + 4)) + ln(e-x(1+ex)) + C
= -4x + 4*ln(ex + 4) - x + ln(ex + 1) + C
= -5x + 4*ln(ex + 4) + ln(ex + 1) + C
I hope none of your other homework questions are as difficult as that was.
Hope this helps, and I apologize for how long this is.