We begin with a u-substitution with #u=ln(x)#. We then divide by the derivative of #u# to integrate with respect to #u#:
#(du)/dx=1/x#
#int\ x^ln(x)\ dx=int\ x*x^u\ du#
Now we need to solve for #x# in terms of #u#:
#u=ln(x)#
#x=e^u#
#int\ x*x^u\ du=int\ e^u*(e^u)^u\ du=int\ e^(u^2+u)\ du#
You might guess that this does not have an elementary anti-derivative, and you'd be right. We can however use the form for the imaginary error function, #erfi(x)#:
#erfi(x)=int\ 2/sqrtpie^(x^2)\ dx#
To get our integral into this form, we may only have one squared variable in the exponent of #e#, so we need to complete the square:
#u^2+u=(u+1/2)^2+k#
#u^2+u=u^2+u+1/4+k#
#k=-1/4#
#u^2+u=(u+1/2)^2-1/4#
#int\ e^(u^2+u)\ du=int\ e^((u+1/2)^2-1/4)\ du=e^(-1/4)int\ e^((u+1/2)^2)\ du#
Now we can introduce a u-substitution with #t=u+1/2#. The derivative is just #1#, so we don't need to do anything special to integrate with respect to #t#:
#e^(-1/4)int\ e^(t^2)\ dt=e^(-1/4)*sqrtpi/2int\ 2/sqrtpie^(t^2)\ dt=e^(-1/4)sqrtpi/2*erfi(t)+C#
Now we can undo all the substitutions to get:
#e^(-1/4)sqrtpi/2erfi(u+1/2)+C=e^(-1/4)sqrtpi/2erfi(ln(x)+1/2)+C#
