Problem 55 - Calculus for Biology and Medicine

Question

Complete the square in the denominator and evaluate the integral. $$ \int \frac{1}{x^{2}-2 x+2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \tan^{-1}(x-1) + C \).

Step 1: Recognize the Standard Form

The given expression in the denominator is of the form \( x^2 - 2x + 2 \). This needs to be converted into a perfect square. The process of completing the square begins by taking the expression \( ax^2 + bx + c \) into the form \((x-h)^2 + k\).

Step 2: Completing the Square

The goal is to express \( x^2 - 2x + 2 \) as a perfect square. Start by focusing on \( x^2 - 2x \). Take half of the coefficient of \( x \), which is -2, giving you -1. Square it to obtain \( (-1)^2 = 1 \). Thus, \( x^2 - 2x = (x-1)^2 - 1 \). Now add 2 to complete the equation as \( (x-1)^2 + 1 \).

Step 3: Substitute Back into Integral

Now that the denominator has been rewritten as a square, substitute this back into the integral. The integral becomes \( \int \frac{1}{(x-1)^2 + 1} \, dx \).

Step 4: Recognize Trigonometric Form

Notice that the form \((x-1)^2 + 1\) matches \(a^2 + u^2\), where \(a = 1\) and \(u = x-1\). This corresponds to the standard form of the arctangent identity integral, \(\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left(\frac{u}{a}\right) + C\).

Step 5: Apply the Integral Result

Apply the formula for the integral. Here, \(a = 1\) and \(u = x-1\), so the integral becomes \(\tan^{-1}(x-1) + C\). Thus, \( \int \frac{1}{(x-1)^2 + 1}\, dx = \tan^{-1}(x-1) + C\).

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to convert a quadratic expression into a form that involves a perfect square. This method is incredibly useful for simplifying integration and solving quadratic equations. When you have a quadratic expression of the form \( ax^2 + bx + c \), the goal is to rewrite it as \((x-h)^2 + k\). Here's a simple approach to complete the square:

• Identify the coefficient \(b\) of the \(x\) term.
• Take half of \(b\), then square it. This value is added and subtracted within the expression to form a perfect square trinomial.
• The expression \(x^2 - 2x + 2\) becomes \((x-1)^2 + 1\) after completing the square.

Completing the square simplifies the complexity of expressions in the denominator, making them easier to integrate, particularly when they align with known integral forms.

Trigonometric Integrals

In calculus, trigonometric integrals involve integrals of functions that include trigonometric functions such as sine, cosine, or tangent. Here, the focus is on expressions that match forms involving trigonometric identities, making the integration process smoother. Suppose you encounter an integral such as:
\[\int \frac{1}{(x-h)^2 + a^2}\, dx\]Notice how it resembles a familiar trigonometric identity?

• Rewriting the expression using algebraic methods like completing the square aligns it with these identities.
• This allows us to recognize it as part of a standard integral formula involving the inverse tangent function.

Understanding these connections is key in integral calculus as it efficiently bridges algebraic manipulation with trigonometric solutions.

Arctangent Function

The arctangent function, denoted as \( \tan^{-1}(x) \), is the inverse of the tangent function. When evaluating integrals that resemble the form \( \int \frac{1}{a^2 + u^2} \, du \), you can use the arctangent identity:
\[\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left(\frac{u}{a}\right) + C\]In the given exercise, after completing the square, you obtain the integral:
\[\int \frac{1}{(x-1)^2 + 1} \, dx\]This matches the form needed to directly apply the arctangent identity by setting \(a = 1\) and \(u = x-1\). From there, the integration becomes straightforward:

• The result is \( \tan^{-1}(x-1) + C \), providing a neat solution using the inverse trigonometric function.

Utilizing this identity simplifies integration tasks and strengthens the understanding of calculus's fascinating interplay between geometry and algebra.