Integral problem: landing this

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The discussion focuses on finding the definite integral of e^(2x)/(e^x + e^(-x)) from 0 to log 2. The integral simplifies to e^x - log(1 + e^(2x)). To evaluate it at log 2, substituting log 2 into the integral yields the expression 2 - log(5). The simplification shows that e^(log 2) equals 2 and e^(2 log 2) equals 4. The final result can be computed further using logarithmic properties or a calculator.
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The problem is to find the definite integral of e^2x/(e^x + e^-x) with the limits of 0 and log 2. Finding the integral was easy enough, e^x - log (1 + e^2x) , but how do I plug log 2 into this. Any hints appreciated.
 
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Hints:

e^{\log x} = x
e^{a\log x} = x^a
 


To plug log 2 into the integral, you can substitute the value of log 2 into the equation in place of x. This will give you the following expression:

e^log 2 - log(1 + e^(2log 2))

Since e^log 2 = 2 and e^(2log 2) = (e^log 2)^2 = 2^2 = 4, the expression simplifies to:

2 - log(1 + 4)

= 2 - log 5

You can then evaluate this expression using a calculator or by using the properties of logarithms to simplify it further. I hope this helps!
 
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