Question

For the following exercises, evaluate the integral using the specified method.\(\int \sqrt{x} \ln (x) d x\) using integration by parts

Short answer

The integral is evaluated as \( \frac{2}{3} x^{3/2} \ln(x) - \frac{4}{9} x^{3/2} + C \).

Step-by-step solution

Identify Parts of Integration by Parts Formula

We begin by identifying parts of the integration by parts formula, which is given as \( \int u \, dv = uv - \int v \, du \). We choose \( u = \ln(x) \) and \( dv = \sqrt{x} \, dx \). The reason is that differentiating \( \ln(x) \) is simpler, and integrating \( \sqrt{x} \) is straightforward.

Differentiate and Integrate

Differentiate \( u \) to find \( du \): \( du = \frac{1}{x} \, dx \). Integrate \( dv \) to find \( v \): \( v = \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \). Now, we have \( u, du, v, \) and \( dv \) to use in the integration by parts formula.

Apply Integration by Parts Formula

Substitute \( u, du, v, \) and \( dv \) into the formula: \[ \int \sqrt{x} \ln(x) \, dx = \ln(x) \cdot \left(\frac{2}{3} x^{3/2}\right) - \int \left(\frac{2}{3} x^{3/2}\right) \cdot \frac{1}{x} \, dx \]. Simplify to get: \[ \frac{2}{3} x^{3/2} \ln(x) - \frac{2}{3} \int x^{1/2} \, dx \].

Solve the Remaining Integral

The remaining integral is \( \int x^{1/2} \, dx \). Integrate it as follows: \( \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \). Substitute this result back into the formula: \[ \frac{2}{3} x^{3/2} \ln(x) - \frac{2}{3} \cdot \frac{2}{3} x^{3/2} \].

Simplify the Expression

Simplify the expression: \[ \frac{2}{3} x^{3/2} \ln(x) - \frac{4}{9} x^{3/2} + C \], where \( C \) is the constant of integration. This is the final evaluation of the integral.

Key concepts

Logarithmic Integration

Logarithmic integration refers to integrals involving logarithmic functions, such as \( \ln(x) \). These kinds of integrals often require special techniques to solve, with integration by parts being one of the most effective methods.
This is because directly integrating logarithmic functions is not straightforward, given their unique properties and the lack of an obvious antiderivative.
In the case of the integral \( \int \sqrt{x} \ln(x) \, dx \), we choose the function \( u \) from integration by parts to be \( \ln(x) \) because differentiating it simplifies the expression to \( du = \frac{1}{x} \, dx \). The simplest expressions often yield more straightforward differentiations, guiding us efficiently through the integration by parts process.

• Integration by parts leverages the Rule: \( \int u \, dv = uv - \int v \, du \). This helps separate terms into easily digestible parts.

By breaking part of the integral into manageable pieces, logarithmic integration becomes a matter of evaluating manageable components rather than complex logs directly.

Differentiation

Differentiation is the process of finding the derivative of a function, giving the function's rate of change at each point.
When applying integration by parts, differentiating is necessary to transform the function \( u \), which is chosen to be a log in this exercise—specifically, \( u = \ln(x) \).
The derivative of \( \ln(x) \) is \( \frac{1}{x} \, dx \), simplifying one element of our integral formula.
This process:

• Makes parts of the original integral easier to evaluate.
• Facilitates calculating the exact change over small intervals, ultimately aiding in integrating over a range.

Differentiation plays a dual role in integration by parts, both for simplifying components and aiding in constructing a new, integrable function. This intertwines with other integral functions to yield a final expression.

Definite Integrals

Definite integrals evaluate the accumulation of quantities over a specific interval, resulting in a number rather than a function, as observed in indefinite integrals.
However, our current problem focuses on the indefinite integral.
Yet, understanding definite integrals is vital, as they employ limits and provide real-world applications like calculating area under curves, which is an extension of what we've discussed here.
An indefinite integral, like our case, remains flexible, evaluated later into definite bounds, or used as a setup for future analysis:

• It represents a family of functions with an arbitrary constant \( C \).
• Passing through specific points (bounds) could become a definite integral, ultimately translating further insights on behaviors of graphical models.

As you gain proficiency with indefinite integrals, transitioning to definite helps relate these calculations to tangible items like distances or total quantities, encompassing full application of calculus techniques.