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April 16, 2014

April 16, 2014

Posted by **D-MO** on Tuesday, March 11, 2008 at 12:33pm.

The correct answer is 3.346, but how?

- calculus -
**Reiny**, Tuesday, March 11, 2008 at 3:07pmv(t) = integral (ln(1+2^t)) + c

to integrate ln(1+2^t) escapes me for the moment. I have tried looking it up in tables, but could not find that pattern in my book of integrals. Online integrators could not handle it either.

Hopefully one of the other tutors could shed some light on this one.

- calculus -
**Count Iblis**, Tuesday, March 11, 2008 at 6:10pmWrite it as:

Log[2^(1/2 t + 1)(2^(-1/2 t) + 2^(1/2 t))/2] =

(1/2 t +1) log(2) +

log[cosh(1/2 log(2) t)]

The integral of log(cosh(p t) can be expressed in terms of Barnes G functions of complex arguments using formula 11 of this site:

http://mathworld.wolfram.com/BarnesG-Function.html

Using a partial integration you can express the integral of log(cosh(p t) in terms of the integral of t tanh(pt) and some simple shifts in the integrastion limits will allow you to find the analytical expression.

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