the integration of (e^x)/(1-E^2x)^3/2 with respect to x
We are currently doing integration by tables, but I can't find the formula that I should use!
Wolfram says:
http://integrals.wolfram.com/index.jsp?expr=e%5Ex%2F%281+-+e%5E%282x%29%29%5E%283%2F2%29&random=false
If u = e^x, then
(e^x)/(1-e^2x)^3/2 dx
= du/(1-u^2)^3/2
Now you can easily find that the integral is
u/√(1-u^2)
which is what Reiny provided.
To find the integration formula for the given expression, we can start by identifying the form of the integrand. In this case, we have
∫ (e^x)/(1 - e^2x)^(3/2) dx.
This is a rational function where the numerator contains an exponential function and the denominator has a square root term. To simplify this expression, we can make a substitution to transform it into a more recognizable form.
Let's substitute u = 1 - e^2x. To find du, differentiate both sides of the equation with respect to x:
du/dx = -2e^2x.
Rearranging this expression, we get dx = -(1/(2e^x))^(-1) du = -(e^x/(2u)) du.
Now, we can rewrite the integral in terms of u:
∫ (e^x)/(1 - e^2x)^(3/2) dx = ∫ (e^x)/(u^(3/2)) (-e^x/(2u)) du = ∫ (-1/2) du.
The last integral is straightforward:
-1/2 ∫ du = -u/2 + C,
where C is the constant of integration.
Finally, undo the substitution by replacing u with its original expression:
-1/2 (1 - e^2x) + C.
Therefore, the final result of the integration is:
∫ (e^x)/(1 - e^2x)^(3/2) dx = -1/2 (1 - e^2x) + C.