Evaluate the integral? : # int x^3e^(x^2) dx #

We need the integration by parts

#intp'qdx=pq-intpq'dx#

We perform the substitution

Let #u=x^2#, #=>#, #du=2xdx#

Therefore,

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

We apply the integration by parts

#p'(u)=e^u#, #=>#, #p(u)=e^u#

#q(u)=u#, #=>#, #q'(u)=1#

Therefore,

#1/2intue^ udu=1/2(ue^u-inte^udu)=1/2(ue^u-e ^u)#

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

We seek:

# I = int \ x^3e^(x^2) \ dx #

Note as a helper that:

# d/dx ( e^(x^2) ) = 2xe^(x^2) iff int \ 2xe^(x^2) \ dx = e^(x^2) #

So we can write the integral as:

# 2I = int \ (x^2)(2xe^(x^2)) \ dx #

We can now use the formula for Integration By Parts (IBP):

# int \ u(dv)/dx \ dx = uv - int \ v(du)/dx \ dx #, or less formally
# " " int \ u \ dv=uv-int \ v \ du #

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Let # { (u,=x^2, => (du)/dx,=2x), ((dv)/dx,=2xe^(x^2), => v,=e^(x^2) ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

gives us

# int \ (x^2)(2xe^(x^2)) \ dx = (x^2)(e^(x^2)) - int \ (e^(x^2))(2x) \ dx #

# :. 2I = x^2e^(x^2) - int \ 2xe^(x^2) \ dx + A #
# " " = x^2e^(x^2) - e^(x^2) + A #
# " " = (x^2-1)e^(x^2) + A #

Hence

# I = 1/2(x^2-1)e^(x^2) + C #