Problem 51 - Calculus Early Transcendentals

Question

Use the reduction formulas in to evaluate the following integrals. $$\int \ln ^{4} x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int \ln^4 x \, dx$$. Answer: The integral $$\int \ln^4 x \, dx$$ evaluates to $$x\ln^4 x - 4x\ln^3 x + 12x\ln^2 x - 24x\ln x + 24x + C$$, where C is the integration constant.

Step 1: Choose u and dv

Choose $$u = \ln^4 x$$ and $$dv = dx$$, so that we can find $$du$$ and $$v$$ next.

Step 2: Find du and v

Differentiate $$u$$ to get $$du = 4\ln^3 x \cdot \frac{1}{x} dx = \frac{4\ln^3 x}{x} dx$$. Integrate $$dv$$ to find $$v = x$$.

Step 3: Apply integration by parts formula

Use the formula $$\int u \, dv = uv - \int v \, du$$, plug in our expressions for $$u, dv, du$$ and $$v$$, then simplify: $$ \int \ln^4 x \, dx = x\ln^4 x - \int x \cdot \frac{4\ln^3 x}{x} dx = x\ln^4 x - 4 \int \ln^3 x \, dx $$

Step 4: Repeat integration by parts

Now, we need to find $$\int \ln^3 x \, dx$$ which can also be evaluated by integration by parts. Let $$u = \ln^3 x$$ and $$dv = dx$$ and find $$du$$ and $$v$$ again. Differentiate $$u$$ to get $$du = 3\ln^2 x \cdot \frac{1}{x} dx = \frac{3\ln^2 x}{x} dx$$ and integrate $$dv$$ to find $$v = x$$. Apply the integration by parts formula again: $$ \int \ln^3 x \, dx = x\ln^3 x - \int x \cdot \frac{3\ln^2 x}{x} dx = x\ln^3 x - 3 \int \ln^2 x \, dx $$

Step 5: Repeat integration by parts again

We repeat the process once more for $$\int \ln^2 x \, dx$$. Let $$u = \ln^2 x$$ and $$dv = dx$$. Differentiate $$u$$ to get $$du = 2\ln x \cdot \frac{1}{x} dx = \frac{2\ln x}{x} dx$$ and integrate $$dv$$ to find $$v = x$$. Apply the integration by parts formula again: $$ \int \ln^2 x \, dx = x\ln^2 x - \int x \cdot \frac{2\ln x}{x} dx = x\ln^2 x - 2 \int \ln x \, dx $$

Step 6: The final integral

The final integral we need to solve is $$\int \ln x \, dx$$. Set $$u = \ln x$$ and $$dv = dx$$, and find $$du$$ and $$v$$ one last time. Differentiate $$u$$ to get $$du = \frac{1}{x}dx$$, and integrate $$dv$$ to find $$v = x$$. Apply the integration by parts formula for the last time: $$ \int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} dx = x\ln x - \int dx = x\ln x - x + C $$ where C is the integration constant. Now, we can substitute this result back into our prior expressions.

Step 7: Combining the results

Combine our results for each integral and substitute them into the original expression to get the final answer: $$ \int \ln^4 x \, dx = x\ln^4 x - 4(x\ln^3 x - 3 \int \ln^2 x \, dx) = x\ln^4 x - 4x\ln^3 x + 12(x\ln^2 x - 2 \int \ln x \, dx) = x\ln^4 x - 4x\ln^3 x + 12x\ln^2 x - 24(x\ln x - x + C) = x\ln^4 x - 4x\ln^3 x + 12x\ln^2 x - 24x\ln x + 24x + C $$ So, the result is: $$ \int \ln^4 x \, dx = x\ln^4 x - 4x\ln^3 x + 12x\ln^2 x - 24x\ln x + 24x + C $$

Key Concepts

Reduction Formulas

Reduction formulas are mathematical expressions that simplify the process of evaluating integrals. They are particularly useful when dealing with complex functions, like powers of logarithmic functions, which are often challenging to integrate directly. These formulas reduce the problem into a series of simpler integrals that can be more easily solved through recursive application.
Reduction formulas follow a specific pattern or recursive relationship, allowing us to express a complex integral in terms of another integral of the same kind but with a reduced degree or order. For example, when integrating powers of logarithmic functions such as \(\int \ln^n x \, dx\), a reduction formula might express it in terms of \(\int \ln^{n-1} x\) integrals. This way, each application decreases the power of the logarithm until the integral becomes straightforward to evaluate.
Using reduction formulas not only simplifies calculations but also provides a systematic method for tackling a wide variety of integral calculus problems. They are particularly beneficial for solving problems that have a repetitive structure, enabling more efficient computation.

Integral Calculus

Integral calculus focuses primarily on the concept of integration, which is the process of finding the integral of a function. Integration is essentially the inverse operation to differentiation and is used to calculate areas, volumes, and other quantities that can be accumulated.
In evaluating integrals, we often use techniques like integration by parts, substitution, or trigonometric identities to simplify the process. Integration by parts, specifically, is crucial when dealing with products of functions, such as in the integral \(\int \ln^n x \, dx\). This method is derived from the product rule for differentiation and is expressed as:
\[ \int u \, dv = uv - \int v \, du \]
By strategically choosing parts of the integrand to differentiate and integrate, we reduce complex integrals into more manageable forms. Throughout the given exercise, integration by parts is repeatedly applied to break down the integral of higher powers of \(\ln x\), making such reductions feasible through a systematic approach.

Logarithmic Functions

Logarithmic functions, denoted as \(\ln(x)\) for the natural logarithm, are inverse functions of exponentials. They play a vital role in integral calculus, particularly due to their unique properties and behaviors.
One distinctive property of logarithms is that logarithmic differentiation can be used to find derivatives of complex functions involving products, quotients, or powers. When it comes to integration, logarithms introduce specific challenges due to their non-linear behavior, but they can be simplified using techniques like integration by parts.
In this exercise, the logarithmic function \(\ln(x)\) is raised to a power, which requires a series of integrations using reduction formulas and integration by parts. It highlights how understanding the characteristics of logarithmic functions is crucial for simplifying and solving integrals with logarithmic terms.
Analyzing logarithmic functions is essential for many areas of mathematics, from solving differential equations to analyzing growth and decay models. Hence, mastering their integration is a valuable skill in integral calculus.