Problem 27 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

The integral of \( \int \frac{x}{\sqrt{x+1}} \, dx \) is \( 2x\sqrt{x+1} - \frac{4}{3}(x+1)^{3/2} + C. \)

Step 1: Choose u and dv

For integration by parts, we select parts of the integral to be \( u \) and \( dv \). Let's choose \( u = x \), so \( du = dx \). Then \( dv = \frac{dx}{\sqrt{x+1}} \) and we need to find \( v \).

Step 2: Find v by Integrating dv

Integrate \( dv = \frac{dx}{\sqrt{x+1}} \). To find \( v \), recognize that this is a substitution integral. Let \( w = x + 1 \), then \( dw = dx \). Then \( v = 2\sqrt{x+1} \), since \( \int w^{-1/2} dw = 2w^{1/2} + C \).

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substituting for \( u \), \( dv \), \( v \), and \( du \) gives:\[ \int \frac{x}{\sqrt{x+1}} \, dx = x(2\sqrt{x+1}) - \int 2\sqrt{x+1} \, dx. \]

Step 4: Simplify and Solve the Remaining Integral

First, simplify the expression:\[ 2x\sqrt{x+1} - 2\int \sqrt{x+1} \, dx. \]Now, integrate \( \int \sqrt{x+1} \, dx \) by using substitution again. Let \( u = x+1 \), then \( du = dx \); thus the integral becomes:\[ 2\int \sqrt{u} \, du = 2 \cdot \frac{2}{3}u^{3/2} = \frac{4}{3}(x+1)^{3/2}. \]

Step 5: Combine All Parts Together

Put it all together:\[ 2x\sqrt{x+1} - \frac{4}{3}(x+1)^{3/2} + C. \] This is the integral of the original function using integration by parts.

Key Concepts

Integral Calculus

Integral calculus is a fascinating branch of mathematics that focuses on the concept of integration. At its core, integration is the process of finding the integral of a function, which often represents the accumulation of quantities. This process is essentially the reverse of differentiation and is a fundamental component in understanding mathematical analysis and its applications. Integral calculus involves two main operations: definite integrals and indefinite integrals. A definite integral gives the total accumulated value between two points, while an indefinite integral represents a family of functions and includes a constant of integration, typically denoted by 'C'. In the exercise above, we worked with an indefinite integral for solving \( \int \frac{x}{\sqrt{x+1}} \, dx \).Understanding integrals helps unlock insights into areas like physics, engineering, and probability, where calculating areas under curves, volumes, and other accumulated changes are essential. Integral calculus not only allows us to solve such real-world problems but also offers a method to reconstruct functions from their rates of change.

Integration Techniques

Integration techniques are various methods used to solve integrals, especially when the function involved is not straightforward. One powerful technique is **Integration by Parts**. It's particularly useful when dealing with the product of two functions. The rule is derived from the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du, \]where

• \( u \) is the first function we're choosing to differentiate
• \( dv \) is the second function part we will integrate

In our example, \( u = x \) whereupon differentiating yields \( du = dx \); and we started with \( dv = \frac{dx}{\sqrt{x+1}} \),a path leading us to further simplify this through the substitution method to find \( v \).Besides integration by parts, there are other techniques such as

• Partial fraction decomposition which splits a complex fraction into simpler parts,
• and Trigonometric substitution for cases dealing with trigonometric expressions.

Selecting the right method is critical and often depends on the form of the integral, calling for a mix of creativity and analytical skills.

Substitution Method

The substitution method, also known as u-substitution, is another common technique in integral calculus, especially for integrals that can be simplified by substituting a part of the integrand with a new variable. This process simplifies the integral into a form that is easier to evaluate.For example, in the integration of \( \int \frac{x}{\sqrt{x+1}} \, dx \), during Step 2,we used the substitution method to deal with \( v = \int \frac{dx}{\sqrt{x+1}} \).By setting \( w = x + 1 \),we transformed our integral into one involving \( \sqrt{w} \),easing its solveability.The general approach for substitution involves:

• Identifying a part of the integrand to replace with a single variable, like substituting \( w = x+1 \) implies \( dw = dx \)
• Rewriting the integral in simpler terms
• Solving the new integral and then substituting back the original variable

The substitution method is especially efficient for solving integrals involving compositions of functions, ensuring smoother and more intuitive calculations.