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September 17, 2014

September 17, 2014

Posted by **abhi** on Thursday, October 4, 2012 at 11:24am.

- calculus - eh? -
**Steve**, Thursday, October 4, 2012 at 11:35amok, what is "e^x 1" ?

- calculus -
**abhi**, Thursday, October 4, 2012 at 12:39pm.Sorry its e^x+1.....:P

- calculus -
**Reiny**, Thursday, October 4, 2012 at 12:46pmIf it is e^(2x)/(e^x + 1) the way you typed it, then the integral is

e^c - ln(e^x + 1) + C

if you meant:

e^(2x)/e^(x+1) then it would simplify to

e^(x-1) and the integral of that is simply

e^(x-1) + c

- calculus -
**abhi**, Thursday, October 4, 2012 at 1:43pmCould u plzz tell it in detail...the 1st 1

- calculus -
**Steve**, Thursday, October 4, 2012 at 2:02pmlet u = e^x+1.

du = e^x dx

and you have

∫(u-1)/u du = ∫ 1 - 1/u du = u - lnu = (e^x+1) - ln(e^x+1) + C

by absorbing the +1 into the C, you end up with

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

- calculus -
**abhi**, Friday, October 5, 2012 at 11:54am...u see the numrtr is e^2x and how can u write it as u-1?

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