Expert: Paul Klarreich - 10/3/2005

Question
Integrate wrt x 2\(e^x +1)

By 1 using the substitution  u = e^x

By 2 multiplying through out by e^-x

By 3 dividing the numerator by the denominator



1   let u = e^x       now   du/dx = e^x   du/e^x = dx
     2/(u+1)du/e^x        but e^x = u
      so 2/(u+1)du/u =  2/u – 2/(u+1)  ( decomposing to partial fractions )

now integrating
= 2lnu- 2ln(u+1) + C


NUMBER 2

= 2e^-x/(1+e^-x) =-2ln(1+e^-x) +  C


3


Answer
Hi, Andrew,
As you can see, the (annoying/delightful) feature of second-semester calculus is the multiple ways a problem can be handled.

[VIEW THIS IN A FIXED-SIZE FONT.]

Integrate wrt x 2\(e^x +1)

By 1 using the substitution u = e^x

By 2 multiplying through out by e^-x

By 3 dividing the numerator by the denominator

1 let u = e^x    now du/dx = e^x du/e^x = dx
   2/(u+1)du/e^x      but e^x = u
   so 2/(u+1)du/u = 2/u – 2/(u+1) ( decomposing to partial fractions )  (YES, THAT SEEMS CORRECT)

now integrating
= 2ln(u)- 2ln(u+1) + C = 2 ln(u/(u+1)) + C, WHICH YOU MAY WANT TO RECONCILE WITH OTHER ANSWERS, THAT MIGHT APPEAR DIFFERENT.  THAT'S THE (BAD PART/FUN PART) OF THESE EXAMPLES.


NUMBER 2

= 2e^-x/(1+e^-x) =-2ln(1+e^-x) + C

You have: (I am using 'I' for the integral symbol.)

I(2e^-x/(1+e^-x)dx, for which you will set u = 1+e^-x
and have du = -e^-x dx.

So you have -2I(du/u) = -2 ln u = as above.

3. Dividing the top by the bottom involves long division.  We all learned this in the fourth grade, but it seems strange to apply it here.  But:

{To simplify the notation, I will write w = e^x

Divide 2\(w +1):

w + 1 ) 2       (2
       2 + w
       ------
         - w, which is the remainder.
       w
= 2 - -----
     w + 1

Now put back w = e^x, and integrate the two terms in much the same manner as the previous examples.

Does that do it?