Difficult integral - how do you do it?

Question

It was a multiple choice, so maybe you can't work it out normally but it was:
[Integral from -1 to 1] (2-e^x)/(1+e^x) dx

a) 0
b) 1
c) ln 2
d) e

My guess is that it will be ln 2 but i'm not sure. How do you do this?

Answer 1

I believe the answer is actually b.
the integration is kinda lengthy, or at least the way I did it, but it does agree with the calculator

let's say u=e^x, du=e^xdx, dx=du/e^x=du/u

Integral[(2-u)/(1+u)(du/u)]=
Integral[(2-u)/(u(1+u))du]

I believe you have to split the rational expression into a sum of fractions. This can be done by starting with:
A/(u) + B/(u+1)=
(Au+A+Bu)/(u(u+1))

now we know that Au + A + Bu =2-u
therefore, A=2 and Au+Bu=-u
so 2u+Bu=-u and B=-3

so the integral is now:
Integral[((2/u) - 3/(u+1))du]
evaluating this we get:
2ln[u]-3ln[u+1]
substituting e^x back in for u
2ln[e^x]-3ln[1+e^x]=
2x-3ln[1+e^x]
evaluate this from x=-1 to x=1
(2-3ln[1+e])-(-2-3ln[1+(1/e)])=
2-3ln[1+e]+2+3ln[(1+e)/(e)]=
4-3ln[1+e]+3ln[1+e]-3ln[e]=
4-3=
1

Answer 2

One trick you can use is long division, the same way you do with polynomials and partial fractions.

(e^x + 1 ) into (-e^x + 2)

Since e^x goes into -e^x -1 times, you'd write a -1 on top, leaving you a remainder of (-e^x + 2) - (-e^x - 1) = 3

So your rearranged integral becomes 3/(e^x + 1) - 1

The -1 is easy to integrate, so we just leave that.

The 3/(e^x + 1) isn't immediately obvious. I'll leave that up to you to solve.

Answer 3

It is b.

int^1_(-1) (2-e^x)/(1+e^x) dx = int^1_(-1) 3/(1+e^x) - 1 dx
= int^1_(-1) 3/(1+e^x) dx -2,

Let's focus on the integral... If we let x=ln(u) then dx=1/u du and we get
int^1_(-1) 2/(1+e^x) dx = int^e_{e^(-1)} 3/(1+u) 1/u du
= int^e_{e^(-1)} 3/u - 3/(1+u) du
= 3 ln(u) - 3 ln(1+u) |^e_{e^(-1)}
= 3 ln(e) - 3 ln(1+e) - 3 ln(e^(-1)) + 3 ln(1+e^(-1))
= 6 - 3ln(1+e) + 3 ln(1+e^(-1))

So the original integral is
6 - ln(1+e) + 3 ln(1+e^(-1)) - 2
= 4 - 3ln(1+e) + 3 ln(1+e^(-1))
= 4 + 3 (ln(1+e)^(-1)+ln(1+e^(-1)))
= 4 + 3 ln( (1+e^(-1))/(1+e))
= 4 + 3 ln(e^(-1)) (Okay, I left a lot of steps out of this... but it's true.)
= 4 -3
which is one.