Problem 13 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int \ln (x+5) d x $$

Step-by-Step Solution

Answer

The integral is \( 5\ln(x+5) - x + C \).

Step 1: Choose a Method

Decide whether to use integration by parts or substitution. For the integral \( \int \ln(x+5) \, dx \), integration by parts is suitable because it deals well with logarithmic functions.

Step 2: Set Up Integration by Parts

Recall the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Here, choose \( u = \ln(x+5) \) and \( dv = dx \).

Step 3: Differentiate and Integrate Parts

Compute \( du \) and \( v \): - Differentiate \( u = \ln(x+5) \): \( du = \frac{1}{x+5} \, dx \) - Integrate \( dv = dx \): \( v = x \).

Step 4: Apply the Integration by Parts Formula

Substitute into the integration by parts formula: \[ \int \ln(x+5) \; dx = x \ln(x+5) - \int x \frac{1}{x+5} \; dx \]

Step 5: Simplify the Remaining Integral

Simplify the expression: \[ \int x \frac{1}{x+5} \, dx = \int \frac{x}{x+5} \, dx \]. Simplify further by rewriting as \( \int \left(1 - \frac{5}{x+5}\right) \, dx = \int 1 \, dx - 5 \int \frac{1}{x+5} \, dx \).

Step 6: Evaluate and Combine the Integrals

Evaluate each part: - \( \int 1 \, dx = x \) - \( \int \frac{1}{x+5} \, dx = \ln|x+5| \) Combine results: \[ \int x \frac{1}{x+5} \, dx = x - 5 \ln(x+5) + C_1 \]

Step 7: Substitute Back Into the Original Expression

Substitute back: \[ \int \ln(x+5) \, dx = x \ln(x+5) - (x - 5 \ln(x+5) + C_1) \]Simplify:\[ = 5 \ln(x+5) - x + C \] where \( C = -C_1 \) is a constant of integration.

Step 8: Verify by Differentiating

Differentiate the solution to check:\[ \frac{d}{dx}[5 \ln(x+5) - x + C] = \frac{5}{x+5} - 1 \]Simplify:\[ = \frac{5 - (x+5)}{x+5} = \frac{-x}{x+5} = \ln(x+5) \] which matches the derivative of \( \int \ln(x+5) dx \).

Key Concepts

Definite Integrals

Definite integrals are an essential concept in calculus, used to find the area under a curve within specified limits. Unlike indefinite integrals, which produce a family of functions, definite integrals result in a real number, representing the accumulated value, such as area or volume.
They are written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. The difference \( b \) and \( a \), represents the interval over which you are integrating.
To solve definite integrals, you:

• Find the antiderivative, which is the function whose derivative is the integrand.
• Calculate the antiderivative at the upper limit \( b \) and lower limit \( a \).
• Subtract the result at \( a \) from the result at \( b \): \( F(b) - F(a) \).

This process gives the net area under the curve from \( a \) to \( b \). It's crucial to remember that if a curve dips below the x-axis, this part contributes a negative area. Thus, the definite integral can sometimes result in a positive, negative, or zero value.

Integration Techniques

When working with integrals, choosing the right integration technique is key to solving the problem efficiently. Here are some common methods:

• Substitution: This technique, similar to the chain rule in differentiation, is useful when the integral involves a function and its derivative. You replace a part of the integral with a single variable, simplifying the integration.
• Integration by Parts: This method comes from the product rule for differentiation. It is used when the integrand is a product of two functions. Choose one function to differentiate and the other to integrate, using the formula: \( \int u \, dv = uv - \int v \, du \).
• Partial Fractions: Applied to rational functions, this technique involves breaking down a complicated fraction into simpler parts that are easier to integrate.

Integration by parts is especially useful when dealing with logarithmic functions, as seen in the exercise. By carefully choosing which part of the integrand to differentiate, you can often simplify complex integrals into more manageable pieces.

Logarithmic Functions

Logarithmic functions, often expressed as \( \ln(x) \), are the inverses of exponential functions. They have interesting properties that make them useful in various mathematical applications:

• Derivatives: The derivative of \( \ln(x) \) is \( \frac{1}{x} \), providing a way to simplify problems involving natural logs during differentiation and integration by parts.
• Logarithm Laws: These include the product, quotient, and power rules, such as \( \ln(ab) = \ln a + \ln b \) and \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \). These laws help in rewriting complex expressions.

In integration, logarithmic functions often require special techniques to integrate. Because their derivative is straightforward, they match well with the integration by parts strategy, as seen with \( \int \ln(x+5) \, dx \) in the exercise. This is because their integration isn't directly evident, requiring manipulation using differentiation of \( u \) and integration of \( dv \) to simplify and solve the integral.