Problem 23 - Applied Calculus

Question

For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals. (a) $$\int \frac{x^{2}}{1+x^{3}} d x$$ (b) $$\int x e^{x^{2}} d x$$ (c) $$\int x^{2} \ln \left(x^{3}+1\right) d x$$ (d) $$\int \frac{1}{\sqrt{3 x+1}} d x$$ (e) $$\int x^{2} \ln x d x$$ (f) $$\int \ln x \, d x$$

Step-by-Step Solution

Answer

(a) Substitution, (b) Substitution, (c) Substitution, (d) Substitution, (e) Parts, (f) Parts.

Step 1: Analyze Integral (a)

For integral \( \int \frac{x^{2}}{1+x^{3}} dx \), notice that the denominator's derivative, \( 3x^2 \), is easily related to the numerator, suggesting the substitution \( u = 1 + x^3 \). Integration by substitution is appropriate.

Step 2: Analyze Integral (b)

For \( \int x e^{x^{2}} dx \), the function \( e^{x^2} \) suggests a substitution where \( u = x^2 \), as the derivative \( 2x \) is present. This makes integration by substitution more suitable.

Step 3: Analyze Integral (c)

In the integral \( \int x^{2} \ln(x^{3}+1) dx \), neither substitution nor parts immediately fits. However, using substitution \( u = x^3 + 1 \), where the derivative \( 3x^2 \) matches a part of the integrand, makes substitution a reasonable choice.

Step 4: Analyze Integral (d)

Looking at \( \int \frac{1}{\sqrt{3x+1}} dx \), setting \( u = 3x + 1 \) allows the square root's derivative \( 1/(2\sqrt{u}) \) to simplify the integrand, making substitution suitable.

Step 5: Analyze Integral (e)

For \( \int x^{2} \ln x \, dx \), the form \( x^n \ln x \) is a classic example for integration by parts, as you can set \( u = \ln x \) and \( dv = x^2 dx \). Thus, integration by parts is more appropriate.

Step 6: Analyze Integral (f)

The integral \( \int \ln x \, dx \) does not lend itself to substitution due to the absence of a clearly related function and its derivative. Using integration by parts, with \( u = \ln x \) and \( dv = dx \), is the standard approach.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique in calculus that simplifies complex integrals by transforming them into an easier form. By choosing a suitable substitution, usually denoted as \( u \), we can convert the integral into a simpler problem. This method is particularly helpful when an integral contains a composite function, where the derivative of the inner function appears somewhere in the integral.

Consider the example from the exercise, integral (a), \( \int \frac{x^{2}}{1+x^{3}} \, dx \). Here, setting \( u = 1 + x^3 \) simplifies the integrand because the derivative of \( u \), \( 3x^2 \, dx \), matches the structure of the numerator when appropriately rescaled. The integral then becomes a simpler function of \( u \) and can be solved more manageably.

Steps to apply this technique effectively include:

• Identify a portion of the integrand as \( u \) such that its derivative is present elsewhere in the integrand.
• Substitute \( u \) and \( du \) into the integral.
• Solve the simplified integral with respect to \( u \).
• Finally, back-substitute to obtain the solution in terms of the original variable.

This approach is useful when you notice that differentiating one part of the function gives you the other part needed to adjust the terms of the integral.

Integration by Parts

Integration by parts is another essential technique, particularly useful when dealing with products of functions like polynomials and logarithms. The method is derived from the product rule for differentiation and is expressed by the formula:\[ \int u \, dv = uv - \int v \, du \]This technique is best applied when you have an integrand that is a product of two functions, and simplifying directly seems difficult.

Let's consider integral (e), \( \int x^{2} \ln x \, dx \). Here, choosing \( u = \ln x \) and \( dv = x^2 \, dx \) is sensible because transforming the logarithmic part using differentiation simplifies the expression, while the polynomial part \( x^2 \) is straightforward to integrate. This swapping of roles between differentiation and integration is pivotal.

Guide to using integration by parts:

• Identify two parts of the integrand, \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to obtain \( v \).
• Apply the integration by parts formula to transform the original integral.
• Simplify and solve the resulting integral, if necessary using the integration by parts method again for any further non-trivial integrals.

Such strategic decomposition of integrals often reveals solutions that are not readily apparent through simple observation.

Calculus Problems

Approaching calculus problems—especially those involving integration—requires a toolkit of strategies to handle different types of integrals. By identifying whether to use substitution or parts, you're essentially breaking down the problem into its basic components and choosing the most efficient path to a solution.

The real skill in solving calculus problems comes with practice and recognizing patterns in the integrals that suggest a specific method. For example, if you see a function under a root or involving exponential expressions, it might hint at substitution like in (b) \( \int x e^{x^2} \, dx \), where setting \( u = x^2 \) simplifies the problem.

For more standard forms like \( \int \ln x \, dx \) in (f), relying on integration by parts can work wonders because substitution does not directly apply. Here, setting \( u = \ln x \) and \( dv = dx \) is a classic move that turns a challenge into a more straightforward solution.

In tackling calculus problems:

• Carefully analyze the form and structure of the integral.
• Choose your integration method based on which part of the integrand is going to simplify the most with differentiation or integration.
• Be ready to iterate between techniques if necessary to get to a final, solvable form.

With practice, these problems not only become easier to solve but also greatly enhance your understanding of the dynamic interactions between different functions in calculus.