How do you find the anti derivative of: [(e^x^2-2x)/(e^x^2)] ?
∫ [(e^x^2-2x)/(e^x^2)]
((The e^x^2 is: e raised to x squared))
if you could explain and show the steps too, thank you
∫ [(e^x^2-2x)/(e^x^2)] dx
= ∫ [1 - 2x/((e^x^2))] dx
= ∫ [1 - 2x(e^(-x^2))] dx
= x - e^(-x^2) + c
check by differentiating
John, how did you get that integral symbol?
Hey, yeah, cool !
I can do a long one
⌠
⌡
but it takes two lines
To find the antiderivative of the given function, we can simplify it first.
Let's rewrite the function as: ∫ [(e^x^2 - 2x) / (e^x^2)] dx.
Now, we can split the numerator into two separate terms: ∫ [e^x^2 / (e^x^2)] dx - ∫ [(2x) / (e^x^2)] dx.
The first term, ∫ [e^x^2 / (e^x^2)] dx, simplifies to ∫ 1 dx, which is just x.
For the second term, ∫ [(2x) / (e^x^2)] dx, we can use u-substitution. Let's set u = x^2, so du = 2x dx. Rearranging, we have dx = du / (2x). Substituting these values back into the integral, we get:
∫ [(2x) / (e^x^2)] dx = ∫ [(2x) / (e^u)] (du / (2x)) = ∫ [1 / (e^u)] du.
Now, this integral is much simpler. The integral of 1 / (e^u) with respect to u is -e^(-u).
Therefore, the antiderivative of ∫ [(e^x^2 - 2x) / (e^x^2)] dx is:
x - (-e^(-x^2)) + C,
where C represents the constant of integration.