Calculus 2 Integration

Question

I'm having trouble understanding this problem.

 

Make the substitution u=e^x to express the integrand as a rational function with three linear factors in the denominator, one of which is u, and then evaluate the integral.

 

∫(-17e^x-20)/(e^2x+5e^x+4)

 

 

Answer 1

Hi Anna.

 

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 = e^x

 

then

 

u² = (e^x)² = e^2x

 

So when you look at your denominator:

 

e^2x + 5e^x + 4

 

what you really have is

 

u² + 5u + 4

 

which you can recognize immediately as

 

(u+4)(u+1)

 

Your numerator is a little more simple

 

-17e^x - 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 = e^x

 

so let's substitute that back in.

 

-16/(e^x + 4) - 1/(e^x + 1)

 

and integrate

 

∫-16/(e^x + 4) - 1/(e^x + 1) dx

 

Factor out e^x from each denominator

 

∫-16/e^x(1 + 4/e^x) - 1/e^x(1 + 1/e^x) dx

 

Do a u-substitution for each term.

 

u1 = 1 + 4/e^x

du1 = -4/e^x dx

-du1 = 4/e^x dx

 

u2 = 1 + 1/e^x

du2 = -1/e^x dx

-du2 = 1/e^x dx

 

Substitute

 

∫4/u1 du1 + ∫1/u2 du2

 

= 4*ln(u1) + ln(u2) + C

 

= 4*ln(1 + 4/e^x) + ln(1 + 1/e^x) + C

 

= 4*ln(e^-x(e^x + 4)) + ln(e^-x(1+e^x)) + C

 

= -4x + 4*ln(e^x + 4) - x + ln(e^x + 1) + C

 

= -5x + 4*ln(e^x + 4) + ln(e^x + 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.

Answer 2

 

∫(-17e^x - 20) / ((e^x)² + 5e^x + 4) dx

 

 

Substituting u=e^x  into the integral, we get

 

∫[(-17u - 20) / (u² + 5u + 4)]du

 

 

Factor the denominator.

 

∫[(-17u - 20) / (u + 4)(u + 1)]du

 

 

From here, you can use partial fractions.  I will give you the initial setup.  Then use that to complete the rest.

 

∫[(-17u - 20) / (u + 4)(u + 1)] du = ∫[ A / (u + 4) + B / (u + 1)] du   --->   setup 1

 

                                                   = ∫[(A(u + 1) + B(u + 4)) / (u + 4)(u + 1)] du  ---> setup 2

 

 

Now you just equate numerators.  From there, equate coefficients to solve for A and B.  Then plug in A and B that you found and find the integral of the sum of the terms in setup 1.  Then once you found your integral, substitute back  u=e^x.