Question

Perform the indicated integrations. $$ \int \frac{d x}{x^{2}+4} $$

Short answer

\[ \int \frac{dx}{x^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \]

Step-by-step solution

Recognize the Form of the Integral

The integral \( \int \frac{dx}{x^2 + 4} \) can be recognized as a standard integral that resembles the form \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). This means we can work towards this form to find the solution.

Identify the Appropriate Formula

In the given integral, \( x^2 + 4 \) corresponds to the standard form \( x^2 + a^2 \) where \( a^2 = 4 \). Therefore, \( a = 2 \). We will use the formula: \[ \int \frac{dx}{x^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \] where \( C \) is the constant of integration.

Apply the Formula

Substitute \( a = 2 \) into the arctangent integration formula: \[ \int \frac{dx}{x^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \]. This gives us the solution for the integral, matching the given function \( x^2 + 4 \).

Verify the Result

Differentiate the result \( \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C \) to check if it gives the original integrand \( \frac{1}{x^2 + 4} \). Use the derivative \( \frac{d}{dx}[\tan^{-1}(u)] = \frac{1}{1+u^2} \cdot \frac{du}{dx} \). For \( u = \frac{x}{2} \), \( \frac{du}{dx} = \frac{1}{2} \). Thus, \[ \frac{d}{dx}\left( \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) \right) = \frac{1}{2} \cdot \frac{1}{1+\left( \frac{x}{2} \right)^2} \cdot \frac{1}{2} = \frac{1}{x^2+4} \]. This confirms the integration was performed correctly.

Key concepts

Arctangent Integration

Understanding arctangent integration is vital for solving integrals where the function resembles the form \(\int \frac{dx}{x^2 + a^2}\). The key insight here is to recognize when you can apply the arctangent formula:\[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C\]In this form, \(a\) is a constant that can be identified from the equation. We used the arctangent integration technique in the solved problem:- First, we identified the form as \(x^2 + 4\) and observed that it aligns with \(x^2 + a^2\).- With \(a^2 = 4\), we found \(a = 2\).By applying this formula, we converted the integral into a calculable expression using arctangent function, including the constant \(C\). Remember, the constant of integration \(C\) accounts for any vertical shifts along the y-axis, representing the family of curves this integration might produce.

Indefinite Integrals

Indefinite integrals are essential in calculus, standing as the counterpart to differentiation. They provide the antiderivative of a function without specific limits, resulting in a general formula.The indefinite integral is represented using the integral symbol without specified integration boundaries:\[\int f(x) \, dx = F(x) + C\]where \(F(x)\) is the antiderivative of \(f(x)\), and \(C\) is the constant of integration. This constant is a crucial part of indefinite integrals since it captures all possible vertical displacements of the function.Key points to remember about indefinite integrals:

• The result is a family of functions parameterized by the constant \(C\).
• No summation limits; hence, it doesn’t yield a numerical value.

In our problem, recognizing the solution as an indefinite integral allowed us to express it as:\[\frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C\]This shows the generalized antiderivative of the given function.

Calculus Techniques

Mastering calculus techniques such as trigonometric substitution and integration by parts is essential in solving complex problems. Trigonometric substitution is particularly valuable for integrals involving terms like \(x^2 + a^2\), as it enables simplification and direct integration using known trigonometric functions.For example, in arctangent integrals, we utilize substitution to address the complexity of certain functions by matching them to standard trigonometric derivatives.Calculus techniques involve:

• Identifying forms of functions that fit known integration models.
• Using algebraic manipulation to simplify integrals.
• Applying trigonometric identities to reduce complexity.

In solving our integral, the key technique was recognizing the integral as an arctangent form and substituting the correct values to apply the formula precisely. This systematic approach illuminates how calculus techniques transform seemingly complicated integration problems into solvable steps.