How do you integrate #int e^(-x)ln 3x dx # using integration by parts?

1 Answer
Feb 5, 2018

#color(blue)( int " " e^(-x)" "ln (3x)" " dx = -e^(-x) ln(3x) - E_1(x)+C#

Explanation:

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#