91Ó°ÊÓ

Completing the square is a method used to simplify quadratic expressions, particularly useful when dealing with integral calculus. It's exactly what the name implies – restructuring a quadratic expression so that it becomes a perfect square trinomial. Let's break it down.

Starting with a general quadratic expression, such as \(x^2 + bx + c\), the goal is to rewrite it in the form \((x-h)^2 + k\). The term \(h\) is determined as \(\frac{b}{2}\), and adding and subtracting \((\frac{b}{2})^2\) helps achieve this structure.

For the example \(x^2+2x+5\), we have:

• The linear coefficient \(b\) is 2, so \(\frac{b}{2} = 1\).
• Add and subtract \(1^2 = 1\), to get: \(x^2 + 2x + 1 - 1 + 5\).
• This rearranges to \((x+1)^2 + 4\).

This transformation helps simplify the integration process by making the expression easier to manipulate or substitute.

Trigonometric substitution is a powerful technique used in integration when dealing with certain types of algebraic expressions. It's particularly useful when an integral involves expressions in the form \(a^2 + x^2\), \(a^2 - x^2\), or \(x^2 - a^2\), as these can be connected to trigonometric identities.

In this exercise, after completing the square, we have an expression \((x+1)^2 + 2^2\). This fits well with the \(a^2 + u^2\) form, where trigonometric identities involving tangent can simplify the problem.

Here's how it simplifies:

• Set the substitution \(u = x + 1\), making \(du = dx\).
• The integral becomes \(\int \frac{1}{u^2 + 2^2} \, du\), aligning with the trigonometric identity.

Trigonometric substitution sets a foundation for the next step in integral calculus, which involves recognizing and applying specific integral formulas.

Arctangent integration is a technique for integrating expressions that match the standard form \(\int \frac{1}{u^2 + a^2} \, du\). This form is particularly well-suited for using arctangent functions, a common function in calculus that helps simplify certain types of integrals.

Here's how it works:

• For \(\int \frac{1}{u^2 + 2^2} \, du\), recognize it corresponds to the standard form \(\int \frac{1}{u^2 + a^2} \, du\).
• Using the formula, the result is \(\frac{1}{a} \tan^{-1} \frac{u}{a} + C\), where \(a\) is 2 in this case.
• This results in \(\frac{1}{2} \tan^{-1} \frac{u}{2} + C\).

After applying the formula, simply reverse the substitution from earlier, replacing \(u\) back with the original expression \(x+1\). This yields the final result: \(\frac{1}{2} \tan^{-1} \frac{x+1}{2} + C\), providing the elegant solution to the integral.