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
