Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int x^{3} \ln x d x $$

Short answer

The indefinite integral of \(x^{3} \ln x dx\) is \(x^{3} \ln x - x^{3} + C\), where C is the constant of integration.

Step-by-step solution

Identify the two functions in the integrand

The two functions in this integral are the polynomial function \(x^{3}\), which will treat as 'u', and the logarithmic function \(\ln x\), which we'll treat as 'v'.

Apply the integration by parts formula

The formula for integration by parts is given by \(\int udv = uv - \int vdu\). We take derivative of function 'u' which gives us \(du = 3x^2 dx\) and integral of function 'v' to get \(dv = 1/x dx\). Substituting \(u\), \(v\), \(du\), \(dv\) into the formula, we get \(\int x^{3} \ln x dx = x^3 \ln x - \int \frac{x^{3}}{x} 3x^2 dx\). Simplifying the right hand side, we get \(x^{3} \ln x - 3\int x^2 dx\). Note that at this point we no longer need to apply integration by parts.

Complete the integration

Now we're ready to do the remaining integral. \(\int x^{2}dx = x^{3}/3\). So, substituting back we get \(\int x^{3} \ln x dx = x^{3} \ln x - 3 * x^{3}/3 = x^{3} \ln x - x^{3}\). Since this is an indefinite integral, don't forget to include the constant of integration in the final answer.

Key concepts

Integration by Parts

The method of integration by parts is a powerful tool in calculus, especially for tackling integrals involving products of functions. The technique is useful when standard methods of integration aren't feasible. It follows the formula:

• \( \int u \ dv = uv - \int v \ du \)

Let's break down this formula into understandable parts.
This formula requires choosing which part of the integrand will be \(u\) (the part to differentiate) and which will be \(dv\) (the part to integrate).
Typically, you set \(u\) as the function that simplifies upon differentiation, while \(dv\) should be something you can easily integrate.
In our example, \( \int x^{3} \ln x \ dx \), the choices are:

• \( u = x^3 \)
• \( dv = \ln x \ dx \)

Differentiating \( u \) gives \( du = 3x^2 \ dx \), and integrating \( dv \) results in \( v = \int \frac{1}{x} dx = \ln x \).
This setup allows us to apply the parts formula and eventually simplifies the problem to an easier integral.

Polynomial Integration

Polynomial integration is one of the more straightforward integrations and serves as a foundation for solving more complex problems.
When dealing with polynomials like \( x^n \), you use the power rule for integration, which states:

• \( \int x^n \ dx = \frac{x^{n+1}}{n+1} + C \ where \ n eq -1 \)

This formula is straightforward and means you add one to the exponent, then divide by the new power.
This method works because the derivative of \( \frac{x^{n+1}}{n+1} \) returns to the original polynomial \( x^n \).
In the context of our problem, after using integration by parts, we land on \( 3 \int x^2 \ dx \).
Using the power rule, \( \int x^2 \ dx = \frac{x^3}{3} \).
This simpler step is crucial in finding the final integral.

Logarithmic Integration

Integrating functions involving logarithms can be a bit tricky compared to polynomial functions.
Understanding the properties and rules of logarithms becomes crucial here.
In our integral \( \int x^{3} \ln x \ dx \), it pairs a polynomial with a logarithm, requiring us to use integration by parts.
Logarithmic functions do not have a straightforward antiderivative like polynomials do.
However, when \( \ln x \) is coupled with other functions, you usually look for ways to simplify or rearrange the integral, which is why integration by parts fits perfectly since it reduces part of the integrand.
Always keep in mind the relationship of logarithms with exponentials and their basic properties, like \(\ln ab = \ln a + \ln b\).
This understanding aids in recognizing patterns or simplifications within an integral that involves logs.