Integral problem: landing this

[redshift]

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.

 

[James R]

Hints:

$$e^{\log x} = x$$
$$e^{a\log x} = x^a$$

 

[wais]

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!