Understanding the Integral Problem 

The question asks us to evaluate the definite integral: ∫e^x {1 + ln x + x ln x}dx. This type of integral often involves recognizing a specific standard form related to the exponential function e^x.

Applying the Standard Integral Formula

There is a useful standard formula for integrals involving e^x:

\(\int e^x \{f(x) + f'(x)\} dx = e^x f(x) + C\)

where \(f'(x)\) is the derivative of \(f(x)\) with respect to \(x\), and \(C\) is the constant of integration.

Rewriting the Integrand

Let's look at the expression inside the integral: \(1 + \ln x + x \ln x\). We need to see if we can separate this expression into a function \(f(x)\) and its derivative \(f'(x)\).

Let's try setting \(f(x) = x \ln x\). Now, let's find the derivative of \(f(x)\) using the product rule \((uv)' = u'v + uv'\):

• Let \(u = x\), then \(u' = \frac{d}{dx}(x) = 1\).
• Let \(v = \ln x\), then \(v' = \frac{d}{dx}(\ln x) = \frac{1}{x}\).

So, the derivative of \(f(x) = x \ln x\) is:

\(f'(x) = \frac{d}{dx}(x \ln x) = u'v + uv' = (1)(\ln x) + (x)\left(\frac{1}{x}\right)\)

\(f'(x) = \ln x + 1\)

Now let's compare this with the original expression inside the integral: \(1 + \ln x + x \ln x\).

We can rewrite the expression as: \(x \ln x + (1 + \ln x)\).

If we let \(f(x) = x \ln x\), then \(f'(x) = 1 + \ln x\).

So, the integrand \(1 + \ln x + x \ln x\) is exactly in the form \(f(x) + f'(x)\), where \(f(x) = x \ln x\) and \(f'(x) = 1 + \ln x\).

Evaluating the Integral

Now that we have identified \(f(x)\) and \(f'(x)\) such that the integral is in the form \(\int e^x \{f(x) + f'(x)\} dx\), we can apply the formula directly:

\(\int e^x \{x \ln x + (1 + \ln x)\} dx\)

Using the formula \(\int e^x \{f(x) + f'(x)\} dx = e^x f(x) + C\), with \(f(x) = x \ln x\), we get:

\(\int e^x \{x \ln x + (1 + \ln x)\} dx = e^x (x \ln x) + C\)

\(= x e^x \ln x + C\)

Final Result

The value of the integral ∫e^x {1 + ln x + x ln x}dx is \(x e^x \ln x + C\). Let's compare this with the given options.

The result \(x e^x \ln x + C\) matches option 1.

Revision Table - Key Concepts

Additional Information - The e^x(f(x)+f'(x)) Formula

The formula \(\int e^x \{f(x) + f'(x)\} dx = e^x f(x) + C\) is derived using integration by parts. Let's briefly look at the derivation:

Consider the integral \(\int e^x f(x) dx\). We can use integration by parts with:

• \(u = f(x) \implies du = f'(x) dx\)
• \(dv = e^x dx \implies v = \int e^x dx = e^x\)

Applying the integration by parts formula ∫ u dv = uv - ∫ v du:

\(\int e^x f(x) dx = f(x) e^x - \int e^x f'(x) dx\)

Now consider the integral we want to evaluate: \(\int e^x \{f(x) + f'(x)\} dx = \int e^x f(x) dx + \int e^x f'(x) dx\).

Substitute the result from the integration by parts step:

\(\int e^x f(x) dx + \int e^x f'(x) dx = (f(x) e^x - \int e^x f'(x) dx) + \int e^x f'(x) dx\)

Notice that the term \(\int e^x f'(x) dx\) cancels out:

\(= f(x) e^x + C\)

This confirms the formula. Recognizing this form in integrals involving e^x can save a lot of time compared to applying integration by parts multiple times.

In our problem, identifying \(f(x) = x \ln x\) and \(f'(x) = 1 + \ln x\) was key to using this shortcut effectively.