Problem 26 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \frac{\ln (x+1)}{\sqrt{x+1}} d x\)

Step-by-Step Solution

Answer

The integral is \( 2 \ln(x+1) \sqrt{x+1} - 4\sqrt{x+1} + C \).

Step 1: Identify the parts

Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \). For the integral \( \int \frac{\ln(x+1)}{\sqrt{x+1}} \, dx \), choose \( u = \ln(x+1) \) and \( dv = \frac{1}{\sqrt{x+1}} \, dx \).

Step 2: Differentiate u

Differentiate \( u = \ln(x+1) \). The derivative is given by \( du = \frac{1}{x+1} \, dx \).

Step 3: Integrate dv

Integrate \( dv = \frac{1}{\sqrt{x+1}} \, dx \). Let \( w = x+1 \), then \( dw = dx \), and the integral becomes \( \int w^{-1/2} \, dw = 2w^{1/2} + C \). Replacing back, we have \( v = 2\sqrt{x+1} + C \).

Step 4: Apply the integration by parts formula

Substitute \( u, du, v, \) and \( dv \) into the integration by parts formula: \( \int \frac{\ln(x+1)}{\sqrt{x+1}} \, dx = \ln(x+1) \cdot 2\sqrt{x+1} - \int 2\sqrt{x+1} \, \frac{1}{x+1} \, dx \). Simplify to \( 2 \ln(x+1) \sqrt{x+1} - \int \frac{2}{\sqrt{x+1}} \, dx \).

Step 5: Simplify the remaining integral

Focus on simplifying \( \int \frac{2}{\sqrt{x+1}} \, dx \). Using substitution again, let \( w = x+1 \), so \( dw = dx \). The integral is \( 2\int w^{-1/2} \, dw = 4\sqrt{w} + C \). Revert \( w \) back to \( x+1 \) to get \( 4\sqrt{x+1} + C \).

Step 6: Compile the final solution

Combine the results to form the final answer: \( \int \frac{\ln(x+1)}{\sqrt{x+1}} \, dx = 2 \ln(x+1) \sqrt{x+1} - 4\sqrt{x+1} + C \).

Key Concepts

Definite Integrals

In calculus, definite integrals are used to calculate the accumulated values over an interval. While indefinite integrals provide a general formula for antiderivatives, definite integrals are more specific. They answer the question: "What is the total accumulation from point A to point B?"

• The notation for definite integrals is usually \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• This calculates the net area under the curve \( f(x) \) from \( x=a \) to \( x=b \).

Understanding definite integrals is essential because they provide the exact value of the quantity over a specific range. In this exercise, although we solve an indefinite integral, knowing how the principles apply when the limits are added, helps deepen the understanding of integration in applied contexts.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change. It helps in determining how a function's value changes with respect to changes in its input.

• The derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
• It represents the slope of the tangent line to the curve at any point \( x \).

In the solution provided, differentiation is used to find \( du \) when \( u = \ln(x+1) \). Computing the derivative leads us to \( du = \frac{1}{x+1} \, dx \), which is crucial for applying the integration by parts technique. This method relies heavily on understanding differentiation as it is necessary to implement the formula correctly.

Mathematical Substitution

Mathematical substitution is a powerful technique used in integration to simplify complex integrals. It involves changing variables to transform the integral into a more manageable form. This approach is particularly useful for functions that can be rewritten through substitutions.

• One common substitution is letting \( w = g(x) \), changing \( dx \) to \( dw \) via \( dw = g'(x) \, dx \).
• After integrating with respect to \( w \), substitute back to the original variable.

In our exercise, substitution is used twice. First, it simplifies \( dv = \frac{1}{\sqrt{x+1}} \, dx \) by letting \( w = x+1 \), and similarly when simplifying \( \int \frac{2}{\sqrt{x+1}} \, dx \). Substitution aids in breaking down intricate integrals into simpler parts, making them easier to solve.