# exponential function integral...

• February 4th 2007, 08:03 PM
thedoge
exponential function integral...
evaluate this integral from [0,1]

int[[(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)]dx]

btw, could someone point me towards the info on how to format these correctly?
• February 4th 2007, 08:10 PM
ThePerfectHacker
Quote:

Originally Posted by thedoge
evaluate this integral from [0,1]

int[[(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)]dx]

btw, could someone point me towards the info on how to format these correctly?

This is a hyperbolic tangent, no?

$\int \tanh 2xdx=\frac{1}{2}\ln (\cosh 2x)+C$
• February 4th 2007, 08:11 PM
Soroban
Hello, thedoge!

Quote:

Evaluate: . $\int^1_0\frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\,dx$

Did you notice that the numerator is almost the derivative of the denominator?

Let $u \:=\:e^{2x} + e^{-2x}\quad\Rightarrow\quad du \:=\:\left(2e^{2x} - 2e^{-2x}\right)\,dx \:=\:2\left(e^{2x} - e^{-2x}\right)\,dx$