Evaluating the Second Integral

Oct 13, 2025 by pynkocean.com 28 views

Introduction

In this article, we will delve into the world of calculus and evaluate the integral of $p^4 \ln p$ . This problem requires the application of integration techniques, specifically the use of substitution and integration by parts. We will break down the solution step by step, making it easy to follow and understand.

Background

The integral in question is $\int p^4 \ln p \, dp$ . This is a definite integral, meaning it has a specific upper and lower bound. However, in this case, we will be evaluating the indefinite integral, which means we will find the antiderivative of the function $p^4 \ln p$ .

Substitution Method

To evaluate the integral, we will use the substitution method. This method involves substituting a new variable into the integral, which simplifies the expression and makes it easier to integrate.

Let's start by substituting $u = \ln p$ . This means that $du = \frac{1}{p} dp$ . We can rewrite the integral in terms of $u$ as follows:

$\int p^4 \ln p \, dp = \int p^4 u \, du$

Integration by Parts

Now that we have substituted $u = \ln p$ , we can use integration by parts to evaluate the integral. Integration by parts is a technique used to integrate products of functions. It states that:

$\int u \, dv = uv - \int v \, du$

In this case, we will let $u = p^4$ and $dv = u \, du$ . This means that $du = 4p^3 dp$ and $v = \frac{1}{4} u^2 = \frac{1}{4} (\ln p)^2$ .

Evaluating the Integral

Now that we have set up the integration by parts, we can evaluate the integral. We will start by finding the antiderivative of $v$ , which is $\frac{1}{4} (\ln p)^2$ . Then, we will multiply this by $u$ , which is $p^4$ . Finally, we will subtract the integral of $v \, du$ , which is $\frac{1}{4} (\ln p)^2 \cdot 4p^3$ .

The antiderivative of $v$ is:

$\frac{1}{4} (\ln p)^2$

Multiplying this by $u$ gives:

$p^4 \cdot \frac{1}{4} (\ln p)^2 = \frac{1}{4} p^4 (\ln p)^2$

Subtracting the integral of $v \, du$ gives:

$\frac{1}{4} p^4 (\ln p)^2 - \int \frac{1}{4} (\ln p)^2 \cdot 4p^3 \, dp$

Simplifying the Integral

Now that we have evaluated the integral, we can simplify it. We will start by simplifying the expression $\frac{1}{4} p^4 (\ln p)^2$ . This can be rewritten as:

$\frac{1}{4} p^4 (\ln p)^2 = \frac{1}{4} p^4 (\ln p)^2$

The integral of $\frac{1}{4} (\ln p)^2 \cdot 4p^3$ is:

$\int \frac{1}{4} (\ln p)^2 \cdot 4p^3 \, dp = \int p^7 \ln p \, dp$

Evaluating the Second Integral

To evaluate the second integral, we will use the same substitution method as before. Let's start by substituting $u = \ln p$ . This means that $du = \frac{1}{p} dp$ . We can rewrite the integral in terms of $u$ as follows:

$\int p^7 \ln p \, dp = \int p^6 u \, du$

Integration by Parts

Now that we have substituted $u = \ln p$ , we can use integration by parts to evaluate the integral. Integration by parts is a technique used to integrate products of functions. It states that:

$\int u \, dv = uv - \int v \, du$

In this case, we will let $u = p^6$ and $dv = u \, du$ . This means that $du = 6p^5 dp$ and $v = \frac{1}{7} u^2 = \frac{1}{7} (\ln p)^2$ .

Evaluating the Second Integral

Now that we have set up the integration by parts, we can evaluate the second integral. We will start by finding the antiderivative of $v$ , which is $\frac{1}{7} (\ln p)^2$ . Then, we will multiply this by $u$ , which is $p^6$ . Finally, we will subtract the integral of $v \, du$ , which is $\frac{1}{7} (\ln p)^2 \cdot 6p^5$ .

The antiderivative of $v$ is:

$\frac{1}{7} (\ln p)^2$

Multiplying this by $u$ gives:

$p^6 \cdot \frac{1}{7} (\ln p)^2 = \frac{1}{7} p^6 (\ln p)^2$

Subtracting the integral of $v \, du$ gives:

$\frac{1}{7} p^6 (\ln p)^2 - \int \frac{1}{7} (\ln p)^2 \cdot 6p^5 \, dp$

Simplifying the Second Integral

Now that we have evaluated the second integral, we can simplify it. We will start by simplifying the expression $\frac{1}{7} p^6 (\ln p)^2$ . This can be rewritten as:

$\frac{1}{7} p^6 (\ln p)^2 = \frac{1}{7} p^6 (\ln p)^2$

The integral of $\frac{1}{7} (\ln p)^2 \cdot 6p^5$ is:

$\int \frac{1}{7} (\ln p)^2 \cdot 6p^5 \, dp = \int p^{11} \ln p \, dp$

Conclusion

In this article, we evaluated the integral of $p^4 \ln p$ . We used the substitution method and integration by parts to simplify the expression and find the antiderivative. The final answer is:

$\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2 + C$

where $C$ is the constant of integration.

Final Answer

The final answer is $\boxed{\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2 + C}$ .

Introduction

In our previous article, we evaluated the integral of $p^4 \ln p$ using the substitution method and integration by parts. In this article, we will answer some common questions related to this problem.

Q: What is the constant of integration?

A: The constant of integration, denoted by $C$ , is a value that is added to the antiderivative to make it equal to the original function. In this case, the constant of integration is $C$ .

Q: How do I evaluate the integral of $p^4 \ln p$ ?

A: To evaluate the integral of $p^4 \ln p$ , you can use the substitution method and integration by parts. First, substitute $u = \ln p$ and $du = \frac{1}{p} dp$ . Then, use integration by parts to evaluate the integral.

Q: What is the antiderivative of $p^4 \ln p$ ?

A: The antiderivative of $p^4 \ln p$ is $\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2 + C$ .

Q: How do I simplify the expression $\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2$ ?

A: To simplify the expression $\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2$ , you can combine like terms and simplify the expression.

Q: What is the final answer to the integral of $p^4 \ln p$ ?

A: The final answer to the integral of $p^4 \ln p$ is $\boxed{\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2 + C}$ .

Q: Can I use other methods to evaluate the integral of $p^4 \ln p$ ?

A: Yes, you can use other methods to evaluate the integral of $p^4 \ln p$ . However, the substitution method and integration by parts are the most common and efficient methods.

Q: What are some common mistakes to avoid when evaluating the integral of $p^4 \ln p$ ?

A: Some common mistakes to avoid when evaluating the integral of $p^4 \ln p$ include:

Not substituting $u = \ln p$ and $du = \frac{1}{p} dp$
Not using integration by parts correctly
Not simplifying the expression correctly
Not adding the constant of integration

Conclusion

In this article, we answered some common questions related to evaluating the integral of $p^4 \ln p$ . We hope that this article has been helpful in clarifying any confusion and providing a better understanding of the problem.

Final Answer

The final answer is $\boxed{\frac{1}{4} p^4 (\ln p)^2 - \frac{1}{7} p^6 (\ln p)^2 + C}$ .