What is the integral of e^2/(3 -e^2)^2 dx from 0 to 1?

Answer 1

Notice that e^2/(3-e^2) is a constant with respect to x. Integral of a constant is just that constant times x.

So now we have that the integral of e^2/(3-e^2)^2 with respect to x from 0 to 1 is [e^2/(e-e^2)^2]*x evaluated from 0 to 1:

[e^2/(e-e^2)^2]*1 - [e^2/(e-e^2)^2]*0 = e^2/(e-e^2)

Answer 2

It's just e²/(3-e²)²)[1-0] = e²/(3-e²)²).

Answer 3

1
∫(e^2/(3 -e^2)^2)dx =
0
(e^2/(3 -e^2)^2) = 0.38357