How do I find this Integral

∫ x².e^(x/2)

Thank you very much!!

Integrate by parts:

u=x^2

dv= e^(x/2) dx

so du = 2x dx

v= 2e^(x/2)

uv-integral(vdu)

x^2* 2e^(x/2) - integral 2e^(x/2) *2x

then you must integrate by parts one more time

leave the first part and concentrate on the integral

u=4x
du = 4
dv= e^(x/2)
v=2e^(x/2)

uv-int(vdu)

4x*2e^(x/2) - int(2e^(x/2)*4)
Now do the final integration
int(8e^(x/2)) = 16e^(x/2)

so total now we have:

x^2* 2e^(x/2) - [4x*2e^(x/2) - 16e^(x/2)]

= x^2 * 2e^(x/2) - 8xe^(x/2) + 16e^(x/2)

∫ x².e^(x/2)
using integration by parts;

let u=x^2.....dv=e^(x/2)
du=2xdx v=1/(1/2)e^(x/2)=2e^(x/2)

∫ uv' dx=uv - ∫u'vdx
=2x^2e^(x/2) - ∫ 2e^(x/2)(2xdx)
=2x^2e^(x/2) - ∫ 4xe^(x/2)dx

let u=x.......dv=e^(x/2)
...du=dx....v=2e^(x/2)

=2x^2e^(x/2) - 4[2xe^(x/2) - ∫ 2e^(x/2)dx
=2x^2e^(x/2) -4[2xe^(x/2)- 4e^(x/2)]+C
=2x^2e^(x/2) - 8xe^(x/2)+16e^(x/2)+C
=2e^(x/2)[x^2- 4x+8]+C answer//

S(x^2)e^(x/2)dx=
2S(x^2)e^(x/2)d(x/2)=
2[(x^2)e^(x/2)-2Sxe^(x/2)dx]=
2[(x^2)e^(x/2)-4Sxe^(x/2)d(x/2)]=
2[(x^2)e^(x/2)-4[xe^(x/2)-Se^(x/2)dx]]…
2(x^2)e^(x/2)-8[xe^(x/2)-2e^(x/2)]=
2(x^2)e^(x/2)-8xe^(x/2)+16e^(x/2)+C