How do you integrate #int x^3*e^(x^2/2) # using integration by parts?

Question

10625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-16T16:01:24

#x^3*e^[x^2/2]*dx=(x^2-2)*e^(x^2/2)+C#

#x^3*e^[x^2/2]*dx#

=#x^2*xe^(x^2/2)*dx#

I used integration by parts with #u=x^2# and #dv=xe^[(x^2/2)]*dx#, so #du=2x*dx# and #v=e^(x^2/2)#

#u*dv=u*v-int v*du#

=#x^2*xe^[(x^2/2)]*dx#

=#x^2*e^(x^2/2)-int 2x*e^(x^2/2)*dx#

=#x^2*e^(x^2/2)-2e^(x^2/2)+C#

=#(x^2-2)*e^(x^2/2)+C#

Answer 2 · 2017-11-16T16:32:59

The answer is #=(x^2-2)e^(x^2/2)+C#

We need

#intpq'=pq-intp'q#

Here,

Perform the substitution

#u=x^2#, #=>#, #du=2xdx#, #=>#, #dx=1/(2x)du#

So,

#intx^3e^(x^2/2)dx=1/2intue^(u/2)du#

#p=u#, #=>#, #p'=1#

#q'=e^(u/2)#, #=>#, #q=2e^(u/2)#

Therefore,

#intx^3e^(x^2/2)dx=1/2intue^(u/2)du#

#=1/2(2ue^(u/2)-2inte^(u/2)du)#

#=1/2(2ue^(u/2)-4e^(u/2))#

#=x^2e^(x^2/2)-2e^(x^2/2)+C#

#=(x^2-2)e^(x^2/2)+C#