Question #1f484

For #intsqrtxe^(-sqrtx)dx#, let #u=sqrtx# and #du=1/2sqrtxdx#

Then #intsqrtxe^(-sqrtx)dx=2intu^2e^udu#

For the integrand #u^2e^(-u) #, integrate by parts, #int f dg = f g - int g df#, where
#f = u^2, dg = e^(-u) du, df = 2 u du, g = -e^(-u): 2intu^2e^udu = -2 e^(-u) u^2 + 4 int ue^(-u) du#

For the integrand #ue^(-u) #, integrate by parts, #int f dg = f g - int g df#, where
#f = u, dg = e^(-u) du, df = du, g = -e^(-u): -2 e^(-u) u^2 + 4 int ue^(-u) du = -4 e^(-u) u - 2 e^(-u) u^2 + 4 int e^(-u) du#

The integral of #e^(-u)# is -#e^(-u):#

#-2 e^(-u) u^2 + 4 int ue^(-u) du= -2 e^(-u) u^2 - 4 e^(-u) u - 4 e^(-u) + c#

Substitute back for #u = sqrt(x): = -2 e^(-sqrt(x)) x - 4 e^(-sqrt(x)) sqrt(x) - 4 e^(-sqrt(x)) + c#

Which is equal to:

Answer: #= -2 e^(-sqrt(x)) (x + 2 sqrt(x) + 2) + c#

We'll start this integration by using substitution.
For #intsqrtx(e^-sqrtx)dx#, let #u=-sqrtx=-x^(1/2)#.
Therefore, #(du)/dx[-x^(1/2)]=-d/dx[x^(1/2)]=-(1/2x^(-1/2))=-1/(2sqrtx)#, so #du=-1/(2sqrtx)dx#.
To substitute we need something that's part of the integrand, so rearrange this to #-2sqrtxdu=dx#

Now our integral looks like #intsqrtx(e^u)(-2sqrtxdu)=int-2xe^udu#.
Remember that #u=-sqrtx#, therefore #u^2=x# so our integral is equivalent to #int-2u^2e^udu#, which we can solve.

First apply the constant rule, noting that #int-2u^2e^udu=-2color(green)(intu^2e^udu)#

Now we need to use integration by parts, which states that #intfdg=fg-intgdf#. Let #f=u^2# and #dg=e^udu#. #(df)/(du)=d/(du)[u^2]=2u#, so #df=2udu#. #intdg=inte^udu=e^u#, so #g=e^u#

#color(green)(intu^2e^udu)=u^2e^u-inte^u(2udu)=u^2e^u-2color(purple)(intue^udu)#

We need to use integration by parts again. This time, let #f=u# and #dg=e^udu#. #(df)/(du)=d/(du)[u]=1#, so #df=du#. #intdg=inte^udu=e^u#, so #g=e^u#

#color(purple)(intue^udu)=ue^u-inte^udu=ue^u-e^u#

#color(green)(intu^2e^udu)=u^2e^u-2color(purple)((ue^u-e^u))=u^2e^u-2ue^u+2e^u#

#-2(color(green)(u^2e^u-2ue^u+e^u))=-2u^2e^u+4ue^u-4e^u#

Finally, plug #-sqrtx# back in for #u# and simplify.

#-2xe^-sqrtx-4sqrtxe^-sqrtx-4e^-sqrtx=-2e^-sqrtx(x+2sqrtx+2)#

We've omitted the constant of integration thus far since all we were doing was integration by parts and substitution, so we'll put it in now.

#-2e^-sqrtx(x+2sqrtx+2)+C#