91Ó°ÊÓ

Completing the square is an essential algebraic technique used to simplify quadratic expressions. This method involves transforming a quadratic expression into perfect square form. Let's see how it works for the expression in our integral, specifically the denominator: \[ x^2 + 4x + 5. \]1. **Identify the Coefficient:** Start by identifying the coefficient of the linear term (the term with 'x'), which is 4 in this case.2. **Divide and Square:** Take this coefficient, divide it by 2, and square the result: \[ \left( \frac{4}{2} \right)^2 = 4. \]3. **Reform the Expression:** Add and subtract this squared value (4) within the expression: \[ x^2 + 4x + 4 - 4 + 5. \] The transformed quadratic can now be expressed as a perfect square: \[ (x+2)^2 + 1. \]This process of completing the square simplifies integration, as it transforms the expression into a recognizable form.

Trigonometric substitution is a method used to evaluate integrals involving quadratic expressions. After completing the square, the integral's denominator becomes a form that matches known trigonometric identities. Our completed square expression \((x+2)^2 + 1\) makes the integral:\[ \int \frac{1}{(x + 2)^2 + 1} \, dx. \]This resembles the trigonometric identity:\( 1 + \tan^2(\theta) = \sec^2(\theta). \)Hence, we can substitute using:- Let \( t = x + 2 \), turning the expression into \( t^2 + 1 \).- Note that \( dt = dx, \) simplifying our substitution further.By using this technique, we have connected the problem to the arctangent function, aiding in seamless integration.

The arctangent function is directly linked to our integral problem. Once we substitute \( t = x + 2 \), the integral:\[ \int \frac{1}{t^2 + 1} \, dt \]can be evaluated using the known result:- The integral of \( \frac{1}{t^2 + 1} \) is \( \arctan(t) + C, \) where \( C \) is the integration constant.On finding the antiderivative of the transformed integral, we substitute back the original variable:\[ \arctan(x + 2) + C. \]This result shows us how trigonometric functions and inverse functions like arctangent frequently appear in calculus problems after using techniques such as completing the square and substitution.