Evaluate The Integral Of Ln(4s+6) Using Integration By Parts

by ADMIN 61 views

Introduction to Integration by Parts

In the realm of calculus, integration by parts is a powerful technique used to evaluate integrals of products of functions. It is derived from the product rule for differentiation and is particularly useful when dealing with integrals involving transcendental functions, such as logarithms and inverse trigonometric functions, multiplied by algebraic functions. This method allows us to transform a complex integral into a simpler one, making it easier to solve. The formula for integration by parts is given by:

∫udv=uv−∫vdu\int u dv = uv - \int v du

Where u and v are functions of a variable (in our case, s), du is the derivative of u, and dv is the derivative of v. The key to successfully applying integration by parts lies in choosing appropriate functions for u and dv. A helpful guideline for this selection is the acronym LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. This acronym suggests the order in which functions should be chosen for u; functions appearing earlier in the list are generally better choices for u.

Before diving into the specifics of evaluating the integral of ln(4s+6), it's crucial to understand the underlying principle of integration by parts. It essentially reverses the product rule of differentiation. When we differentiate a product of two functions, say u(s)v(s), we get:

dds[u(s)v(s)]=u′(s)v(s)+u(s)v′(s)\frac{d}{ds} [u(s)v(s)] = u'(s)v(s) + u(s)v'(s)

Integrating both sides with respect to s gives us:

∫dds[u(s)v(s)]ds=∫[u′(s)v(s)+u(s)v′(s)]ds\int \frac{d}{ds} [u(s)v(s)] ds = \int [u'(s)v(s) + u(s)v'(s)] ds

u(s)v(s)=∫u′(s)v(s)ds+∫u(s)v′(s)dsu(s)v(s) = \int u'(s)v(s) ds + \int u(s)v'(s) ds

Rearranging the terms, we arrive at the integration by parts formula:

∫u(s)v′(s)ds=u(s)v(s)−∫v(s)u′(s)ds\int u(s)v'(s) ds = u(s)v(s) - \int v(s)u'(s) ds

This formula is the cornerstone of integration by parts and allows us to shift the integration from one part of the product to another, potentially simplifying the integral. In the context of our problem, we will see how choosing u and dv strategically can lead us to a solution for the integral of ln(4s+6).

Problem Statement: Integral of ln(4s+6)

The core of this article focuses on evaluating the integral: $\int \ln (4 s+6) d s$. This integral presents a unique challenge because it involves a logarithmic function, which doesn't have a straightforward antiderivative. To tackle this, we employ the technique of integration by parts, a method particularly suited for integrals involving products of functions or, in this case, a single function that can be treated as a product with the constant function 1. The natural logarithm function, ln(4s+6), is a classic example where integration by parts shines. It allows us to transform the integral into a form that is easier to solve by strategically choosing parts of the integrand to differentiate and integrate.

The integral ∫ln(4s+6) ds falls under the category where integration by parts is not just a useful tool but often the most effective, if not the only, method to find a solution. The presence of the natural logarithm, a transcendental function, suggests that setting it as the 'u' part in the integration by parts formula is a wise approach. This is because differentiating a logarithmic function simplifies it, potentially making the resulting integral less complex. On the other hand, we need a 'dv' part, and in this case, the implied '1' (since we can think of the integral as ∫ln(4s+6) * 1 ds) becomes our choice. Integrating '1' is straightforward, giving us 's', which is manageable.

Understanding the problem statement is the first step towards solving it. We recognize that we're dealing with the integral of a logarithmic function, and direct integration is not immediately obvious. This prompts us to consider alternative techniques, with integration by parts being the most promising. The next step involves carefully applying the integration by parts formula, choosing appropriate functions for 'u' and 'dv', and then solving the resulting integral. The challenge here is not just in applying the formula but in making the right choices that simplify the problem, turning a seemingly complex integral into a solvable one. The application of integration by parts in this context highlights its power and versatility in handling integrals that are not amenable to basic integration rules.

Applying Integration by Parts

To effectively evaluate the integral $\int \ln (4 s+6) d s$ using integration by parts, we must first identify the appropriate functions for u and dv. Following the LIATE rule, where Logarithmic functions are prioritized, we choose:

u=ln⁡(4s+6)u = \ln(4s + 6)

dv=dsdv = ds

This choice is strategic because differentiating u will simplify the logarithmic function, while integrating dv is straightforward. Now, we find du and v:

du=dds[ln⁡(4s+6)]ds=44s+6dsdu = \frac{d}{ds} [\ln(4s + 6)] ds = \frac{4}{4s + 6} ds

v=∫dv=∫ds=sv = \int dv = \int ds = s

With u, dv, du, and v determined, we can apply the integration by parts formula:

∫udv=uv−∫vdu\int u dv = uv - \int v du

Substituting our chosen functions, we get:

∫ln⁡(4s+6)ds=sln⁡(4s+6)−∫s⋅44s+6ds\int \ln(4s + 6) ds = s \ln(4s + 6) - \int s \cdot \frac{4}{4s + 6} ds

The original integral is now transformed into a new integral, $\int s \cdot \frac{4}{4s + 6} ds$, which we need to evaluate. This new integral is a rational function, and while it might seem daunting at first, it is generally simpler to handle than the original logarithmic integral. The key to successfully applying integration by parts lies in this transformation – shifting the complexity from the original integrand to a more manageable form.

The next step involves simplifying and solving the new integral. This often requires algebraic manipulation or further integration techniques. In this case, we'll focus on simplifying the rational function inside the integral to make it easier to integrate. The goal is to break down the fraction into simpler terms, possibly using techniques like partial fraction decomposition or algebraic manipulation. The success of integration by parts hinges on this step – the ability to transform a complex integral into a simpler, solvable one. The strategic choice of u and dv sets the stage, but the subsequent steps determine the final outcome. Thus, the initial application of integration by parts is only the first act in the play of solving the integral; the subsequent simplification and solution of the resulting integral are equally crucial.

Simplifying the Resulting Integral

After applying integration by parts, we arrived at the integral:

∫ln⁡(4s+6)ds=sln⁡(4s+6)−∫s⋅44s+6ds\int \ln(4s + 6) ds = s \ln(4s + 6) - \int s \cdot \frac{4}{4s + 6} ds

The focus now shifts to simplifying and evaluating the integral $\int s \cdot \frac{4}{4s + 6} ds$. First, we can simplify the constant by bringing it outside the integral:

4∫s4s+6ds4 \int \frac{s}{4s + 6} ds

To further simplify the integrand, we can use a simple algebraic manipulation. We add and subtract a constant in the numerator to make it a multiple of the denominator. Notice that if we add and subtract 6/4 = 3/2 in the numerator, we can rewrite the fraction:

s4s+6=14⋅4s4s+6\frac{s}{4s + 6} = \frac{1}{4} \cdot \frac{4s}{4s + 6}

Now, add and subtract 6 from the numerator:

14⋅4s+6−64s+6=14(1−64s+6)\frac{1}{4} \cdot \frac{4s + 6 - 6}{4s + 6} = \frac{1}{4} \left(1 - \frac{6}{4s + 6}\right)

So, the integral becomes:

4∫s4s+6ds=4∫14(1−64s+6)ds4 \int \frac{s}{4s + 6} ds = 4 \int \frac{1}{4} \left(1 - \frac{6}{4s + 6}\right) ds

=∫(1−64s+6)ds= \int \left(1 - \frac{6}{4s + 6}\right) ds

This simplification makes the integral much easier to solve. We have transformed a rational function into a difference of two terms, one of which is a constant and the other is a simpler rational function. This step is crucial because it breaks down a complex integral into more manageable components. The algebraic manipulation performed here highlights the importance of recognizing patterns and applying techniques to simplify integrands before attempting direct integration. The success of this step paves the way for a straightforward integration process, bringing us closer to the final solution of the original integral.

Solving the Simplified Integral

Having simplified the integral, we now have:

∫(1−64s+6)ds\int \left(1 - \frac{6}{4s + 6}\right) ds

This integral can be easily solved by applying the basic rules of integration. We can split the integral into two separate integrals:

∫1ds−∫64s+6ds\int 1 ds - \int \frac{6}{4s + 6} ds

The first integral, $\int 1 ds$, is straightforward:

∫1ds=s\int 1 ds = s

For the second integral, $\int \frac{6}{4s + 6} ds$, we can use a simple substitution. Let:

u=4s+6u = 4s + 6

du=4dsdu = 4 ds

ds=14duds = \frac{1}{4} du

Substituting these into the integral, we get:

∫64s+6ds=6∫1u⋅14du=64∫1udu\int \frac{6}{4s + 6} ds = 6 \int \frac{1}{u} \cdot \frac{1}{4} du = \frac{6}{4} \int \frac{1}{u} du

The integral of $\frac{1}{u}$ is a standard result:

64∫1udu=32ln⁡∣u∣\frac{6}{4} \int \frac{1}{u} du = \frac{3}{2} \ln|u|

Substituting back for u, we have:

32ln⁡∣4s+6∣\frac{3}{2} \ln|4s + 6|

Now, combining the results of the two integrals, we get:

∫(1−64s+6)ds=s−32ln⁡∣4s+6∣\int \left(1 - \frac{6}{4s + 6}\right) ds = s - \frac{3}{2} \ln|4s + 6|

This step is the culmination of the simplification process. By breaking down the integral into simpler parts and applying basic integration rules and substitution, we have successfully evaluated the integral. The solution obtained here is a critical component of the final solution to the original problem. The ability to recognize standard integral forms and apply appropriate techniques, such as substitution, is essential in solving integrals efficiently. The successful integration of the simplified form demonstrates the power of strategic simplification and the importance of mastering fundamental integration techniques.

Final Solution

Now that we have evaluated the simplified integral, we can substitute it back into the integration by parts equation to obtain the final solution. Recall that we had:

∫ln⁡(4s+6)ds=sln⁡(4s+6)−∫s⋅44s+6ds\int \ln(4s + 6) ds = s \ln(4s + 6) - \int s \cdot \frac{4}{4s + 6} ds

And we found that:

∫s⋅44s+6ds=s−32ln⁡∣4s+6∣\int s \cdot \frac{4}{4s + 6} ds = s - \frac{3}{2} \ln|4s + 6|

Substituting this back into the equation, we get:

∫ln⁡(4s+6)ds=sln⁡(4s+6)−(s−32ln⁡∣4s+6∣)+C\int \ln(4s + 6) ds = s \ln(4s + 6) - \left(s - \frac{3}{2} \ln|4s + 6|\right) + C

Where C is the constant of integration. Finally, we can simplify the expression:

∫ln⁡(4s+6)ds=sln⁡(4s+6)−s+32ln⁡∣4s+6∣+C\int \ln(4s + 6) ds = s \ln(4s + 6) - s + \frac{3}{2} \ln|4s + 6| + C

This is the final solution to the integral. It combines the results of the integration by parts and the subsequent simplification and integration of the resulting integral. The constant of integration, C, is a crucial part of the final answer, as it represents the family of antiderivatives for the given function. The solution obtained here demonstrates the effectiveness of the integration by parts technique in handling integrals involving logarithmic functions. The step-by-step process, from choosing u and dv to simplifying and solving the resulting integral, highlights the importance of a systematic approach to problem-solving in calculus. The final solution not only provides the antiderivative of the given function but also showcases the power and versatility of integration techniques in calculus.

Conclusion

In conclusion, we have successfully evaluated the integral $\int \ln(4s + 6) ds$ using the integration by parts technique. This method is particularly useful for integrals involving products of functions, and in this case, it allowed us to handle the integral of a logarithmic function. The key steps in the process were:

  1. Choosing appropriate functions for u and dv based on the LIATE rule.
  2. Applying the integration by parts formula to transform the integral.
  3. Simplifying the resulting integral using algebraic manipulation.
  4. Solving the simplified integral using basic integration rules and substitution.
  5. Combining the results to obtain the final solution, including the constant of integration.

This exercise demonstrates the power and versatility of integration by parts as a tool in calculus. It also highlights the importance of strategic problem-solving, including recognizing when to apply specific techniques and how to simplify complex expressions. The final solution, $s \ln(4s + 6) - s + \frac{3}{2} \ln|4s + 6| + C$, provides a complete antiderivative for the given function, showcasing the effectiveness of the method in finding integrals that are not immediately obvious. The process of evaluating this integral serves as a valuable example of how integration by parts can be applied to solve a wide range of problems in calculus and beyond. The mastery of such techniques is essential for anyone studying mathematics, physics, engineering, or any field that relies on calculus as a fundamental tool.