Problem 55 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int \frac{d x}{2 \sqrt{x}+2 x} $$

Step-by-Step Solution

Answer

\( \ln |1 + \sqrt{x}| + C \)

Step 1: Simplify the Integrand

Start by factoring out a common factor from the denominator. The integrand is \( \frac{1}{2\sqrt{x} + 2x} \). Factor out 2 from the denominator:\[\int \frac{dx}{2(\sqrt{x} + x)} = \int \frac{1}{2} \cdot \frac{dx}{\sqrt{x} + x}.\]

Step 2: Substitute for Simplicity

Use substitution to simplify the expression. Let \( u = \sqrt{x} \), then \( x = u^2 \) and \( dx = 2u \, du \). The integral becomes:\[\int \frac{1}{2} \cdot \frac{2u \, du}{u + u^2} = \int \frac{u \, du}{u + u^2}.\] Simplify the integrand by factoring the denominator:\[\int \frac{u \, du}{u(1 + u)}.\] Now, the expression reduces to:\[\int \frac{du}{1 + u}.\]

Step 3: Integrate the Reduced Expression

The integral \( \int \frac{du}{1 + u} \) can be solved using a standard formula:\[\int \frac{du}{1+u} = \ln |1+u| + C,\] where \(C\) is the constant of integration.

Step 4: Substitute Back the Original Variable

Convert the expression back to the variable \(x\). Recall that \( u = \sqrt{x} \), so:\[\ln |1 + u| = \ln |1 + \sqrt{x}|.\] Therefore, the original integral evaluates to:\[\ln |1 + \sqrt{x}| + C.\]

Key Concepts

Substitution Method

The substitution method is a powerful technique used in integration to simplify the process of finding an antiderivative. This method is particularly useful when dealing with complex integrals that involve compositions of functions.
By substituting a part of the integrand with a new variable, the integral's complexity is reduced, making it easier to solve.
This works well with integrals where a function and its derivative appear together. To use substitution:

• Identify a part of the integral to substitute, often an inner function or the root expression.
• Express that part in terms of a new variable, say, let \( u = g(x) \).
• Differentiate \( u \) to find \( du \), often \( du = g'(x) dx \), and rearrange for \( dx \).
• Replace all instances of the identified part with \( u \) and \( dx \) with the expression in terms of \( du \).

This reduces the integral into an easier form that can be integrated directly.
Don't forget to convert back to the original variable after integration.

Integration by Parts

Integration by parts is another essential technique in integral calculus. It stems from the product rule for derivatives and is beneficial for integrating the product of two functions.
It can be expressed through the formula: \[\int u \, dv = uv - \int v \, du\] Here is how to apply it:

• Select \( u \) and \( dv \) from the original integral \( \int u \, dv \), where \( dv \) is the part that can be easily integrated.
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to obtain \( v \).
• Substitute these results into the integration by parts formula.
• Solve the resulting integrals as necessary.

This method is especially useful in cases where differentiation simplifies a function, and integration of \( dv \) is straightforward.
However, it might sometimes need to be iterative, applying the formula multiple times to fully integrate the expression.

Definite Integrals

Definite integrals provide the area under a curve between two specific limits or bounds. This is opposed to indefinite integrals, which represent a family of functions, not a specific value.
The notation for a definite integral is: \[\int_{a}^{b} f(x) \, dx\] Where \( a \) and \( b \) are the lower and upper limits, respectively. The steps to evaluate a definite integral involve:

• Integrating the function \( f(x) \) as if it were an indefinite integral, finding its antiderivative \( F(x) \).
• Then, use the Fundamental Theorem of Calculus to evaluate this antiderivative at the upper limit and subtract the evaluation at the lower limit: \( F(b) - F(a) \).

Definite integrals are powerful in applications requiring exact values, such as calculating total distances, areas, and quantities over a specified range.

Indefinite Integrals

Indefinite integrals, or antiderivatives, refer to a family of functions that differ by a constant.
In mathematical terms, it is represented by: \[\int f(x) \, dx = F(x) + C\] Where \( F(x) \) is the antiderivative of \( f(x) \), and \( C \) is the constant of integration, which accounts for all potential shifts in the function along the vertical axis. When dealing with indefinite integrals:

• Identify the function \( f(x) \) you need to integrate.
• Apply integration techniques such as substitution or integration by parts to find an antiderivative.
• Add the constant \( C \) to represent the family of solutions.

Indefinite integrals are crucial in problems where only the form of the function is essential, providing foundational solutions for further analysis or application.