Step 1: Understanding the Concept:
The problem asks to evaluate an indefinite integral of a rational function where the denominator is a quadratic expression.
When the denominator is a quadratic that cannot be factored easily into real linear factors (i.e., its discriminant is negative), we use the method of "completing the square."
Completing the square allows us to transform the quadratic into a form that resembles the derivative of the inverse tangent function (\( \arctan \)).
Step 2: Key Formula or Approach:
The standard integral form for inverse tangent is:
\[ \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
We need to manipulate the denominator \( x^2 + 4x + 5 \) into the form \( (x+h)^2 + a^2 \).
Step 3: Detailed Explanation:
Consider the quadratic expression in the denominator: \( x^2 + 4x + 5 \).
To complete the square:
1. Look at the coefficient of \( x \), which is \( 4 \).
2. Divide it by \( 2 \) to get \( 2 \), and then square it to get \( 4 \).
3. Rewrite the constant term \( 5 \) as \( 4 + 1 \).
\[ x^2 + 4x + 5 = (x^2 + 4x + 4) + 1 \]
\[ = (x + 2)^2 + 1^2 \]
Now, substitute this expression back into the integral:
\[ I = \int \frac{1}{(x + 2)^2 + 1^2} dx \]
We can use a simple substitution: Let \( u = x + 2 \), then \( du = dx \).
The integral becomes:
\[ I = \int \frac{1}{u^2 + 1^2} du \]
Using the standard formula \( \int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C \) where \( a = 1 \):
\[ I = \frac{1}{1} \tan^{-1} \left( \frac{u}{1} \right) + C \]
\[ I = \tan^{-1}(u) + C \]
Replace \( u \) with the original expression \( x + 2 \):
\[ I = \tan^{-1}(x + 2) + C \]
Step 4: Final Answer:
The result of the integration is \( \tan^{-1}(x + 2) + C \).
This is exactly what is shown in Option (A).