Given:
#color(brown)(int " "e^xln(x) " "dx)# ... Expression.1
Integration by Parts: #color(green)(int " " f*g' = fg - int" "f'g)#
We will integrate by parts
Referring to our problem, we have
#color(brown)(f=ln(x) and g = e^x" "#
While solving our problem,
we will consider the following known results in Calculus:
#color(brown)(f' = 1/x and g' = e^x)# and
#color(brown)(int " "e^x dx = e^x + C)#
Now, we can write our Expression.1 as
#ln(x)*e^x - int " "[1/x*e^x]" " dx#
#rArr e^x*ln(x) - int " "e^x/x * dx# ... Expression.2
#color(brown)(--------------------)#
Note:
#color(blue)(Ei(x)# represents the exponential integral
#color(green)(int_(-oo)^x x^t/t dt)#
For #color(red)(x>0,# the integral #color(red)(Ei(x)# is interpreted as Cauchy Principal Value
#color(brown)(--------------------)#
We will use the above note on #color(red)(Ei(x))# in writing our final solution
We rewrite our ... Expression.2 as
#color(blue)(e^x*ln(x)-Ei(x)+C)#
Hope you find this solution useful.
