Question

a. Evaluate \(\int x \ln x^{2} d x\) using the substitution \(u=x^{2}\) and evaluating \(\int \ln u d u\). b. Evaluate \(\int x \ln x^{2} d x\) using integration by parts. c. Verify that your answers to parts (a) and (b) are consistent.

Short answer

Question: Evaluate the integral \(\int x \ln{x^2}~dx\) using both substitution method and integration by parts, and verify the answers are consistent. Solution: Using substitution method, we found that \(\int x \ln{x^2}~dx = \frac{1}{2}(x^2(\ln{x^2} - 1) + C)\) Using integration by parts, we found that \(\int x \ln{x^2}~dx = x^2(\ln{x^2} - 1) + C\) Since the results from both methods are consistent, the integral \(\int x \ln{x^2}~dx\) is equal to \(\frac{1}{2}(x^2(\ln{x^2} - 1) + C)\) or \(x^2(\ln{x^2} - 1) + C\).

Step-by-step solution

Part (a): Substitution Method

Let's first perform a substitution by letting \(u=x^2\). Now, we compute the differential of \(u\) with respect to \(x\): \begin{align*} \frac{d u}{d x} = 2x \end{align*} So, \(d u = 2x ~d x\). Now we substitute \(u\) and \(d u\) into the initial integral: \begin{align*} \int x \ln{x^2}~d x &= \int \ln{u} \cdot \frac{1}{2x} ~d u \end{align*} This simplifies to: \begin{align*} \frac{1}{2} \int \ln{u} ~d u \end{align*} Now, we need to integrate \(\ln u\) with respect to \(u\).

Part (a): Integration of \(\ln u\)

To integrate \(\ln u\) with respect to \(u\), we will use integration by parts in this step. Let: \(v = \ln u\) and \(dw = du\). Then, we find the derivatives and integrals: \begin{align*} dv &= \frac{1}{u} du \\ w &= u \end{align*} Now, we apply integration by parts: \begin{align*} \int v dw &= vw - \int w dv \\ &= u \ln u - \int u \cdot \frac{1}{u} du \\ &= u \ln u - \int du \end{align*} Integrating both sides, we have \begin{align*} \int \ln u du &= u (\ln u - 1) + C \end{align*} So, our simplified integral becomes \begin{align*} \frac{1}{2} \int \ln u du &= \frac{1}{2}(u(\ln u - 1) + C) \end{align*} Now, we substitute back \(u = x^2\) to get the final answer: \begin{align*} \int x \ln{x^2}~d x &= \frac{1}{2}(x^2(\ln{x^2} - 1) + C) \end{align*}

Part (b): Integration by Parts

Now, let's solve the original integral using integration by parts. Let: \(v = x\) and \(dw = \ln{x^2} ~d x\). Then, we find the derivatives and integrals: \begin{align*} dv &= d x \\ w &= \int \ln{x^2} ~d x \end{align*} We need to find the integral of \(\ln{x^2}\) to find \(w\). We use the property \(\ln{a^b} = b \ln{a}\): \begin{align*} w &= \int 2\ln x ~d x \end{align*} Now, using integration by parts again (with \(a = \ln x\) and \(db = dx\)) \begin{align*} da &= \frac{1}{x} dx \\ b &= x \end{align*} Now, applying integration by parts, \begin{align*} \int a db &= ab - \int b da \\ &= x\ln x - \int x \cdot \frac{1}{x} dx \\ &= x\ln x - \int dx \end{align*} We left with: \begin{align*} w &= 2(x\ln x - x + C) \end{align*} Now, we can apply integration by parts to our original integral: \begin{align*} \int v d w &= vw - \int w dv \\ &= x(2(x \ln x - x + C)) - \int (2(x \ln x - x + C)) dx \end{align*} Solve the remaining integral, and we get \begin{align*} \int x \ln{x^2}~d x &= x^2(\ln{x^2} - 1) + C \end{align*}

Part (c): Verification

Now we compare the results from parts (a) and (b) to verify that they are consistent: \begin{align*} \frac{1}{2}(x^2(\ln{x^2} - 1) + C) &= x^2(\ln{x^2} - 1) + C \end{align*} Both expressions are equal, so the results from parts (a) and (b) are consistent.

Key concepts

Substitution Method

The substitution method in integration is a practical tool that simplifies complex integrals. By changing the variable, we can turn a difficult problem into a simpler one, making it easier to solve. This is achieved by substituting part of the integral with a new variable, then doing the calculus based on the new variable.
In the exercise, the integral \( \int x \ln x^{2} \, dx \) can be simplified by using the substitution \( u = x^2 \). This way, the integral becomes \( \int \ln u \cdot \frac{1}{2x} \, du \).

• The derivative of \( u = x^2 \) is \( du = 2x \, dx \).
• By inserting this into the equation, it allows us to proceed with our solution more conveniently.

The result leads us to simplify the integral significantly, helping us solve it faster and more efficiently.

Integration by Parts

Integration by parts is another vital technique that assists in solving integrals that involve a product of functions. It is derived from the product rule of differentiation and allows for transforming the integral of a product of functions into simpler forms. The formula is:
\[ \int u \, dv = uv - \int v \, du \]
In our problem, we used integration by parts to solve \( \int x \ln x^{2} \, dx \). In this case:

• Let \( u = x \) and \( dv = \ln x^2 \, dx \).
• This provides us with \( du = dx \) and \( v = \int 2 \ln x \, dx \).

At this point, using integration by parts allows evaluation of the integral, breaking it down into finds for \( uv \) and \( \int v \, du \).
This method essentially transforms challenging integrals into sums and differences of simpler ones, making computation easier.

Natural Logarithm Integration

Integrating the natural logarithm function can often require clever techniques, as seen in our question. The integral \( \int \ln u \, du \) doesn't have a simple antiderivative, but it can be integrated using integration by parts.
We set \( v = \ln u \) and \( dw = du \), which gives:

• \( dv = \frac{1}{u} \, du \)
• \( w = u \)

Applying the integration by parts formula here results in:
\[ \int \ln u \, du = u(\ln u - 1) + C \]
This relationship simplifies the task when faced with the integration of logarithmic functions in calculus. It is crucial to master this, because natural logarithms are common in real-world applications and combine easily with other functions.