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

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: $\int \text{ " e^(-x)" "ln (3x)" } \mathrm{dx} ,$

color(red)(f = ln(3x) and g'=(e^-x)

color(green)(Step.1

We will differentiate:

color(red)(d/(dx)[ ln(3x)]

$\Rightarrow \left[\frac{1}{3 x}\right] \cdot \frac{d}{\mathrm{dx}} \left[3 x\right]$

$\Rightarrow \frac{3 \cdot \frac{d}{\mathrm{dx}} \left[x\right]}{3 x}$

$\Rightarrow \frac{1}{x}$

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

$\Rightarrow \mathrm{dx} = - \mathrm{du}$

$\Rightarrow - \int \text{ } {e}^{u} \cdot \mathrm{du}$

$\Rightarrow - {e}^{u}$

$\Rightarrow - {e}^{- x}$ Substitute back $u = - x$

$\therefore \int \text{ } {e}^{- x} \cdot \mathrm{dx} = - {e}^{- x} + C$

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:

$\Rightarrow - \int - \frac{{e}^{- x}}{x} \mathrm{dx} \cdot - {e}^{- x} \cdot \ln \left(3 x\right)$

Note that,

$\int \text{ "-(e^(-x))/x " } \mathrm{dx}$

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