Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int t^{5} \ln \left(t^{7}\right) d t $$

Short answer

The integral evaluates to \( \frac{t^6}{6} ( \ln(t^7) - \frac{7}{6} ) + C \).

Step-by-step solution

Understand the Problem

We are given the integral \( \int t^5 \ln(t^7) \, dt \) and are asked to solve it using integration by parts, which is a technique. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \).

Choose \( u \) and \( dv \)

For this integral, let us choose \( u = \ln(t^7) \), which makes its derivative simpler. The differential of \( u \) is \( du = \frac{7}{t} \, dt \). We set \( dv = t^5 \, dt \), and upon integrating, we get \( v = \frac{t^6}{6} \).

Apply Integration by Parts Formula

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we substitute what we found: \[ uv = \ln(t^7) \cdot \frac{t^6}{6} \] and \[ v \, du = \int \frac{t^6}{6} \cdot \frac{7}{t} \, dt = \frac{7}{6} \int t^5 \, dt \].

Simplify and Integrate

Substitute back into the integration by parts formula: \[ \int t^5 \ln(t^7) \, dt = \ln(t^7) \cdot \frac{t^6}{6} - \frac{7}{6} \int t^5 \, dt \]. Now, simplify the remaining integral \( \int t^5 \, dt = \frac{t^6}{6} \).

Final Integration Result

Putting it all together, the integral becomes: \[ \frac{t^6 \ln(t^7)}{6} - \frac{7}{6} \cdot \frac{t^6}{6} + C \]. Simplify to get: \[ \frac{t^6}{6} ( \ln(t^7) - \frac{7}{6} ) + C \], where \( C \) is the constant of integration.

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus. It is used to calculate the exact area under a curve, between two points on the x-axis. When we compute a definite integral, we're working with limits of integration, typically denoted as \[ \int_{a}^{b} f(x) \, dx \] where \( f(x) \) is the function we're integrating, and \( a \) and \( b \) are the bounds.
The result of a definite integral gives a numeric value, representing the accumulated quantity (like area or total change) over the interval \([a, b]\). We solve definite integrals by finding the antiderivative, evaluating it at the upper and lower limits, and subtracting:
- Evaluate the antiderivative at \( b \)- Subtract the antiderivative evaluated at \( a \)- Therefore, \[ F(b) - F(a) \]
This technique is a powerful tool for solving a wide variety of problems in physics and engineering where one needs to accurately measure physical quantities. Understanding definite integrals is essential to mastering the field of calculus.

Antiderivative

An antiderivative represents the reverse process of differentiation in calculus. When we find an antiderivative, we're essentially determining a function whose derivative is the original function.
This is often referred to as finding the 'indefinite integral'. For a function \( f(x) \), an antiderivative is a function \( F(x) \) such that:\[ F'(x) = f(x) \]
One of the key aspects of antiderivatives is the constant of integration, denoted as \( C \). This constant arises because differentiation of a constant gives zero, implying there are infinitely many antiderivatives for a given function. Hence, any antiderivative can be expressed as:
- \( F(x) + C \)
Antiderivatives are crucial in solving integration problems and in finding definite integrals. They allow us to reconstruct a function from its rate of change, which is fundamental in many scientific observations and calculations.

Logarithmic Functions

Logarithmic functions play a pivotal role in integration, especially when using techniques like integration by parts. Understanding that a logarithm is the inverse of exponentiation informs us about their properties and how they are manipulated. The most common logarithmic function is the natural logarithm:
- \( \ln(x) \)
Logarithmic identities can simplify integration processes. For instance, the derivative of \( \ln(x) \) is:\[ \frac{d}{dx} \ln(x) = \frac{1}{x} \]This derivative property makes logarithmic functions good candidates for integration by parts, since they typically simplify expressions.
In the given exercise, we utilize the \( \ln(t^7) \) as part of the integration by parts step. Breaking it down, \( \ln(t^7) = 7 \ln(t) \), which utilizes the power rule of logarithms:
- \( \ln(a^b) = b \ln(a) \)
Getting familiar with these functions and their properties is integral to performing more complex calculus operations.

Calculus Techniques

Calculus features a range of techniques used to solve complex integrals and derivatives. Among these, integration by parts often stands out when faced with challenging integration problems. This is based on the product rule for differentiation.
Integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]
We pick parts of the integrand to differentiate and integrate to simplify the original integral. It's often helpful when one part of a product is easily integrable, and the other when differentiated, simplifies the problem. This was applied in the provided exercise where \( u \) was selected as a logarithmic function due to its simpler derivative.
Other calculus techniques often used alongside are:

• Substitution – simplifying an integral by changing variables
• Partial fractions – breaking down complex rational functions
• Trigonometric identities – simplifying expressions involving sin, cos, etc.

Mastering these methods enhances problem-solving capabilities, equipping students for more advanced topics and real-world applications.