calculus
posted by DMO .
A Particle moves along the xaxis so that at any time t>0, its acceleration is given by a(t)= ln(1+2^t). If the velocity of the particle is 2 at time t=1, then the velocity of the particle at time t=2 is?
The correct answer is 3.346, but how?

v(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. 
Write 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/BarnesGFunction.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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