Question

use integration by parts to derive the given formula. $$ \begin{array}{rl} \int x^{\alpha}(\ln x)^{2} & d x=\frac{x^{\alpha+1}}{\alpha+1}(\ln x)^{2} \\ \- & 2 \frac{x^{\alpha+1}}{(\alpha+1)^{2}} \ln x+2 \frac{x^{\alpha+1}}{(\alpha+1)^{3}}+C, \alpha \neq-1 \end{array} $$

Short answer

Use integration by parts twice on \(\int x^\alpha (\ln x)^2 dx\) to derive the formula by simplifying each step.

Step-by-step solution

Select Functions for Integration by Parts

In integration by parts, we use the formula \( \int u\, dv = uv - \int v\, du \). Let's select \( u = (\ln x)^2 \) and \( dv = x^\alpha dx \). Our goal is to choose \( u \) such that its derivative simplifies the integrand.

Differentiate and Integrate Selected Functions

Now, differentiate \( u \) and integrate \( dv \). For \( u = (\ln x)^2 \), we have \( du = 2(\ln x) \cdot \frac{1}{x} dx = \frac{2 \ln x}{x} dx \). For \( dv = x^\alpha dx \), we integrate to get \( v = \frac{x^{\alpha+1}}{\alpha+1} \).

Apply Integration by Parts Formula

Using the integration by parts formula, substitute \( u \), \( dv \), \( du \), and \( v \) to find \( \int u\, dv \). We have: \[ \int x^\alpha (\ln x)^2 dx = \frac{x^{\alpha+1}}{\alpha+1} (\ln x)^2 - \int \frac{x^{\alpha+1}}{\alpha+1} \cdot \frac{2 \ln x}{x} dx. \]

Simplify the Integral Result

The integral simplifies to: \[ \frac{x^{\alpha+1}}{\alpha+1} (\ln x)^2 - \frac{2}{\alpha+1} \int x^{\alpha} \ln x \, dx. \] The next step is to evaluate \( \int x^{\alpha} \ln x \, dx \) using integration by parts again, since this is a case of another integral that requires the same technique.

Apply Integration by Parts Again

To solve \( \int x^{\alpha} \ln x \, dx \), set \( u = \ln x \) and \( dv = x^\alpha dx \). Then \( du = \frac{1}{x} dx \) and \( v = \frac{x^{\alpha+1}}{\alpha+1} \). Using integration by parts, it results in: \[ \int x^{\alpha} \ln x \, dx = \frac{x^{\alpha+1}}{\alpha+1} \ln x - \int \frac{x^{\alpha+1}}{\alpha+1} \cdot \frac{1}{x} dx. \]

Simplify the Second Integral

Simplify and solve \( \int \frac{x^{\alpha+1}}{x} dx = \int x^{\alpha} dx \), which gives \( \frac{x^{\alpha+1}}{\alpha+1} + C_1 \). Substitute this back to get: \[ \int x^{\alpha} \ln x \, dx = \frac{x^{\alpha+1}}{\alpha+1} \ln x - \frac{x^{\alpha+1}}{(\alpha+1)^2} + C_1. \]

Substitute Back to Original Equation

Substitute the result from Step 6 into the expression from Step 4: \[ \int x^\alpha (\ln x)^2 dx = \frac{x^{\alpha+1}}{\alpha+1} (\ln x)^2 - \frac{2}{\alpha+1} \left( \frac{x^{\alpha+1}}{\alpha+1} \ln x - \frac{x^{\alpha+1}}{(\alpha+1)^2} \right) + C. \]

Simplify and Verify the Final Expression

Simplify the above expression to reach the given result: \[ \frac{x^{\alpha+1}}{\alpha+1} (\ln x)^2 - 2 \frac{x^{\alpha+1}}{(\alpha+1)^2} \, \ln x + 2 \frac{x^{\alpha+1}}{(\alpha+1)^3} + C. \] Ensure each term aligns with the original formula.

Key concepts

Integral Calculus

Integral calculus is a foundational part of mathematics that deals with the integration of functions. It is the counterpart to differential calculus, where the focus is finding the derivative, or the rate of change of a function. In integral calculus, the primary goal is to determine the total accumulation of quantities, such as areas under curves. This is achieved through the process known as integration. Integration allows us to find important values like areas, volumes, central points, among other things.

There are various methods to integrate different kinds of functions, and integration by parts is a popular technique used. The integration by parts formula is derived from the product rule of differentiation and is given by:

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

This formula is particularly useful when an integral is a product of two functions. Selecting which function to differentiate (\( u \)) and which to integrate (\( dv \)) can dramatically simplify solving an integral.

Definite Integrals

While indefinite integrals have an arbitrary constant \( C \), definite integrals calculate the exact accumulation between two specific limits. The format for a definite integral is:

• \( \int_{a}^{b} f(x) \, dx \)

where \( a \) and \( b \) are the lower and upper boundaries, respectively.

The definite integral finds real numbers that represent the exact area under the curve of a function between these boundaries. This is crucial in applications ranging from physics to engineering, as it allows precise calculations, such as determining the total distance traveled from a velocity function.

Integration by parts is also applicable to definite integrals. The only difference comes in the evaluation step, where you need to calculate the integrated terms at the boundary values and subtract the lower from the upper values:

• \( \left[ uv \right]_a^b = u(b)v(b) - u(a)v(a) \)

This approach is integral in fields that require precise boundary calculations.

Logarithmic Functions

Logarithmic functions have the form \( \log_b(x) \), generally with base 10 or the natural base \( e \) (termed the natural logarithm \( \ln(x) \)). In calculus, logarithmic functions are often encountered in integration due to their derivative and integral properties.

The natural logarithm function \( \ln(x) \) is especially important. Its derivative \( \frac{d}{dx} \ln x = \frac{1}{x} \) and its integral \( \int \frac{1}{x} dx = \ln |x| + C \) are both widely used in solving calculus problems. Moreover, when dealing with products involving logarithms, integration by parts can simplify the process significantly by reducing the power of a logarithm, as seen with expressions like \( (\ln x)^2 \).

In the original exercise, the integral \( \int x^{\alpha} (\ln x)^2 \, dx \) emphasizes the complexity that arises when both algebraic and logarithmic functions coexist. The strategy used in solving this involves setting \( u = (\ln x)^2 \), whereby the complexity of the logarithmic term is reduced step by step. This leverages the decreasing nature of high power logarithmic terms when differentiated, simplifying calculations.