We have the integral
#int_0^1x^2(x^3+1)^3dx#
To deal with this without expanding #(x^3+1)^3# and then integrating term by term, which would be slow, there's a more elegant solution that involves substitution:
Let #u=x^3+1#, which implies that #du=3x^2dx#. Since we have only #x^2dx# in the integrand and not #3x^2dx#, multiply the integrand by #3# and the exterior of the integral by #1/3#.
#=1/3int_0^1(x^3+1)^3 3x^2dx#
Now, we can substitute #u# and #du#. However, in doing this, we need to change the bounds of the integral by plugging the current bounds into #u=x^3+1#. The bound of #0# becomes #0^3+1=1# and the bound of #1# becomes #1^3+1=2#.
#=1/3int_1^2u^3du#
Integration of #u^3# yields #u^4/4+C#, so we can evaluate the integral using that antiderivative:
#=1/3[u^4/4]_1^2=1/3(2^4/4-1^4/4)=1/3(4-1/4)=1/3(15/4)=5/4#
