How would I find the integral of int (x*e^-x^2)?

Question

I used parts where u = e^-x^2 and dv = x but I cannot seem to get it. It keeps going on forever. I know the answer is -e^-x^2/2 from the Integrator. I tried using the method to get identical integrals on the left and divide it by 2 but I can never get that.

x * e^(-x^2)

Thanks.

Answer 1

I suppose youn mean Integral x e^(-x^2) dx. Just observe that the derivative of x^2 is 2x. So, this integral can be written as

-1/2 Integral e^(-x2) (-2x) dx. Since -2x is the derivative of -x^2, we now have an integral of the kind of

Integral e^u du = e^u + C

And our answer is, therefore,

Integral x e^(-x^2) dx = (-1/2) Integral e^(-x2) (-2x) dx. = -1/2 e^(-x^2) + C = -(e^(-x^2)) + C.

Answer 2

∫ ( x e^(-x^2) dx )

This is solved using substitution and not integration by parts. Let me demonstrate by rearranging the integral; this intermediate step will show the substitution in action.

∫ (e^(-x^2) x dx )

Let u = -x^2. Then
du = -2x dx. From here, we multiply both sides by (-1/2) to get
(-1/2)du = x dx

Note that x dx is the tail end of our integral, so after the substitution, (-1/2) du will be the tail end. With that said, applying the substitution, we get

∫ (e^u (-1/2) du )

Factor out the (-1/2) from the integral to get

(-1/2) ∫ ( e^u du )

And now we have an easy integral to solve.

(-1/2)e^u + C

But u = -x^2, so our final answer is

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

Answer 3

Use the substitution y=x^2. Then dy=2*x*dx, so int (x*e^-x^2) becomes
int (1/2*e^-y dy). Peform this easy integration then reverse the substitution and remember to add the arbitrary constant.

Answer 4

you should use this 'trick', x^2=t, and then 2*xdx=dt, which in turn gives xdx=dt/2, so you just replace e^-x^2 with e^-t and xdx with dt/2, so your new integral is (1/2)*e^-t*dt and integral of that is -(1/2)*e^-t, and when you replace back t for x^2 you'll get -(1/2)*e^-x^2.

Answer 5

e^(-x^2)/2