Solving the Integral of \((x^x)^2 (1 + \ln x)\)

The question asks us to find the value of the integral \( \int (x^x)^2 (1 + \ln x) dx \). To solve this integral, we first simplify the expression inside the integral sign and then look for a suitable method of integration.

Understanding the Integral Expression

The term \((x^x)^2\) can be simplified using the exponent rule \((a^b)^c = a^{b \cdot c}\). So, \((x^x)^2 = x^{x \cdot 2} = x^{2x}\).

The integral now becomes:

\(\int x^{2x} (1 + \ln x) dx\)

This form suggests that the integrand might be the result of differentiating a function involving \(x^{2x}\) multiplied by a constant.

Using Differentiation to Solve the Integral

Let's consider the derivative of a function of the form \(f(x)^{g(x)}\). Specifically, let's differentiate \(x^{2x}\) with respect to \(x\). We can use logarithmic differentiation for this.

Finding the Derivative of \(x^{2x}\)

Let \(y = x^{2x}\). To differentiate this, we take the natural logarithm of both sides:

\(\ln y = \ln(x^{2x})\)

Using the logarithm property \(\ln(a^b) = b \ln a\), we get:

\(\ln y = 2x \ln x\)

Now, differentiate both sides with respect to \(x\). Remember to use the chain rule on the left side and the product rule on the right side (\(\frac{d}{dx}(uv) = u'v + uv'\)).

The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y} \frac{dy}{dx}\).

The derivative of \(2x \ln x\) with respect to \(x\) is:

\(\frac{d}{dx}(2x \ln x) = \frac{d}{dx}(2x) \cdot \ln x + 2x \cdot \frac{d}{dx}(\ln x)\)

\(= 2 \cdot \ln x + 2x \cdot \frac{1}{x}\)

\(= 2 \ln x + 2\)

So, equating the derivatives:

\(\frac{1}{y} \frac{dy}{dx} = 2 \ln x + 2\)

Now, multiply by \(y\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = y (2 \ln x + 2)\)

Substitute \(y = x^{2x}\) back:

\(\frac{dy}{dx} = x^{2x} (2 \ln x + 2)\)

Factor out 2 from the term in the parenthesis:

\(\frac{dy}{dx} = 2 x^{2x} (\ln x + 1)\)

This is the derivative of \(x^{2x}\). Notice that the term \(x^{2x} (1 + \ln x)\) in our integral is exactly half of this derivative.

Completing the Integration

We found that the derivative of \(x^{2x}\) is \(2 x^{2x} (1 + \ln x)\). This means that \(x^{2x} (1 + \ln x) = \frac{1}{2} \frac{d}{dx}(x^{2x})\).

Now we can rewrite the integral:

\(\int x^{2x} (1 + \ln x) dx = \int \frac{1}{2} \frac{d}{dx}(x^{2x}) dx\)

Using the property that the integral of the derivative of a function is the function itself (plus a constant of integration):

\(\int \frac{d}{dx}(f(x)) dx = f(x) + C\)

In our case, \(f(x) = x^{2x}\). So, the integral is:

\(\int \frac{1}{2} \frac{d}{dx}(x^{2x}) dx = \frac{1}{2} \int \frac{d}{dx}(x^{2x}) dx\)

\(= \frac{1}{2} x^{2x} + C\)

where \(C\) is the constant of integration.

Comparing Solution with Options

Our calculated integral is \(\frac{1}{2} x^{2x} + C\). Let's compare this with the given options:

1. x ^2x + c
2. \(\frac{1}{2}\) x^2x + c
3. 2x ^2x + c
4. \(\frac{1}{2}\) x^x + c

Comparing our result with the options, we see that it matches option 2.

Revision Table: Calculus Techniques

Additional Information: Logarithmic Differentiation

Logarithmic differentiation is a powerful technique particularly useful when dealing with functions where both the base and the exponent are functions of \(x\), i.e., functions of the form \(y = [f(x)]^{g(x)}\). It simplifies the differentiation process by converting the exponential form into a product using logarithms, which can then be differentiated using the product rule.

The general steps are:

1. Start with the function \(y = f(x)^{g(x)}\).
2. Take the natural logarithm of both sides: \(\ln y = \ln([f(x)]^{g(x)})\).
3. Use the logarithm property \(\ln(a^b) = b \ln a\) to rewrite the right side: \(\ln y = g(x) \ln(f(x))\).
4. Differentiate both sides with respect to \(x\). The left side becomes \(\frac{1}{y} \frac{dy}{dx}\) by the chain rule. The right side is differentiated using the product rule.
5. Solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\), and substitute back the original expression for \(y\).

This technique is also helpful for differentiating complex expressions involving products and quotients of many functions by turning them into sums and differences of logarithms.