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.