First, we divide the integrate in two parts , that we have called I_1 and I_2:
I_1 = intxdx
I_2 = intxe^(4-2x)dx
So F(x) will be:
F(x) = I_1 - I_2
I_1 = intxdx = (x^2)/2 + C_1
I_2 = intxe^(4-2x)dx
With this integral, we need to use the integration by parts theorem, so we define u and v as:
u = x and its derivative u' = dx
dv = e^(4-2x)dx and its integrate v = -1/2e^(4-2x)
So, I_2 is:
I_2 = -x(e^(4-2x))/2 + 1/2inte^(4-2x)dx
And then:
I_2 = -x(e^(4-2x))/2 + 1/2(-1/2)e^(4-2x) + C_2
I_2 = -(e^(4-2x))/2(x+1/2) + C_2
We have to subtract both integrals:
F(x) = I_1 - I_2 = x^2/2 + C_1 + (e^(4-2x))/2(x+1/2) - C_2
C_1 and C_2 are constants so we can group together in one constant that we can define as:
C = C_1 - C_2
So, the final expression of F(x) is:
F(x) = x^2/2 + (e^(4-2x))/2(x+1/2) + C
- Calculation of the constant C
For this, we use the value of F(x) in x=0: F(0) = 1
F(0) = 0 + e^4/2(0 + 1/2) + C = 1
e^4/4 + C = 1
So, the value of C is:
C = 1 - e^4/4