Given:
#color(green)( int " " e^(-x)" "ln (3x)" " dx#
Integration by Parts Method must be used to solve the problem.
The formula is
#color(brown)(int f*g' = f*g-int f'*g#
For our problem: # int " " e^(-x)" "ln (3x)" " dx,#
#color(red)(f = ln(3x) and g'=(e^-x)#
#color(green)(Step.1#
We will differentiate:
#color(red)(d/(dx)[ ln(3x)]#
#rArr [1/(3x)]*d/(dx)[3x]#
#rArr (3*d/(dx)[x])/(3x)#
#rArr 1/x#
#color(brown)[:.d/(dx)[ ln(3x)] = 1/x#
#color(green)(Step.2#
We will next integrate #color(red)(e^(-x)*dx#
#color(red)(int " "e^(-x)*dx#
Substitute #color(green)(u = -x#
#rArr dx = -du#
#rArr -int " "e^u* du#
#rArr -e^u#
#rArr -e^(-x)# Substitute back #u = -x#
#:. int " "e^(-x)*dx = -e^(-x)+C#
#color(green)(Step.3#
We are Given:
#color(green)( int " " e^(-x)" "ln (3x)" " dx#
Refer to the formula
#color(brown)(int f*g' = f*g-int f'*g#
Now we can write our final solution as:
#rArr -int-(e^(-x))/xdx*-e^(-x)*ln(3x)#
Note that,
#int " "-(e^(-x))/x " "dx#
is a special integral and is also an exponential integral
This can be written as #color(red)(E_1 (x)#
Hence, our final solution can be re-written as
#color(blue)( int " " e^(-x)" "ln (3x)" " dx = -e^(-x) ln(3x) - E_1(x)+C#
