∫(e^x - e^-x)/(e^x + e^-x) dx?

Question

How do you solve it?

Answer 1

Well e^x - e^(-x) = d(e^x + e^(-x))/dx So this is set up for substitution

Let u = e^x + e^(-x) Thus du = (e^x - e^(-x)) dx

So ∫(e^x - e^-x)/(e^x + e^-x) dx = ∫du/u

= ln u + C

= ln (e^x + e^-x) + C

Answer 2

Oneo f the ways to solve this problem is to expand it out and solve for each term individually. What we have is a factored difference of squares: (a - b) (a + b) = a^2 - b^2. So we expand it just like that.

Integral ( [e^x]^2 - [e^(-x)]^2) dx. Whenever we have a power to a power, we multiply the exponents; therefore, we get

Integral ( [e^(2x) - e^(-2x) ]) dx

Now, these are relatively simple integrals to solve, because the inside of the function is linear. We *could* use substitution to solve it, but why?
The integral of sin(2x) is -cos(2x)/2. The integral of cos(6x) is sin(6x)/6. All we have to do is divide by the value to offset the chain rule when taking the derivative. Therefore, for

Integral ( [e^(2x) - e^(-2x) ]) dx, we get

[e^(2x)]/2 - [e^(-2x)]/(-2) + C, or

(1/2)e^(2x) + (1/2)(e^(-2x)) + C

This can be simplified further, but I think I'll stop here.

Answer 3

∫tanh x dx=ln cosh x

Answer 4

Note that the numerator is the derivative of the denominator.
That is derivative of e^x is e^x
and derivative of e^(-x) is - e^(-x)

The integral of u' / u, where u is any funtion,
is ln( abs(u) )

Thus the ∫(e^x - e^-x)/(e^x + e^-x) dx

is ln ( /(e^x + e^-x) )

Answer 5

sinh x = e^x - e^(-x)
cosh x = e^x + e^(-x)
tanh x = sinh x / cosh x
integral {tanh x dx} = ln(cosh x) + C
= ln {e^x + e^(-x)} + C

Answer 6

wow... that looks like greek to me. MAJOR props to anyone who can figure that out... or even atempt it lol

=]