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

Given:

Integration by Parts Method must be used to solve the problem.

The formula is

We will differentiate:

We are Given:

Refer to the formula

Now we can write our final solution as:

Note that,

is a special integral and is also an exponential integral

Hence, our final solution can be re-written as

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To integrate ( \int e^{-x} \ln(3x) , dx ) using integration by parts, we use the formula:

[ \int u , dv = uv - \int v , du ]

Let's choose ( u = \ln(3x) ) and ( dv = e^{-x} , dx ). Then, we find the differentials:

[ du = \frac{1}{x} , dx ] [ v = -e^{-x} ]

Now, we apply the integration by parts formula:

[ \int e^{-x} \ln(3x) , dx = -\ln(3x) \cdot e^{-x} - \int -e^{-x} \cdot \frac{1}{x} , dx ]

Simplifying and integrating the second term on the right-hand side, we get:

[ \int e^{-x} \ln(3x) , dx = -\ln(3x) \cdot e^{-x} + \int \frac{e^{-x}}{x} , dx ]

The integral ( \int \frac{e^{-x}}{x} , dx ) does not have a simple elementary form, so we express it in terms of the exponential integral function:

[ \int \frac{e^{-x}}{x} , dx = -\text{Ei}(-x) + C ]

Therefore, the final result of the integral ( \int e^{-x} \ln(3x) , dx ) using integration by parts is:

[ -\ln(3x) \cdot e^{-x} - (-\text{Ei}(-x)) + C ] [ = -\ln(3x) \cdot e^{-x} + \text{Ei}(-x) + C ]

Where ( C ) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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