Question

Perform the indicated integrations. $$ \int \frac{1}{x^{2}-4 x+9} d x $$

Short answer

\( \frac{1}{\sqrt{5}} \arctan\left( \frac{x-2}{\sqrt{5}} \right) + C \).

Step-by-step solution

Identify the integration type

The expression \( \frac{1}{x^2 - 4x + 9} \) suggests the potential use of partial fraction decomposition or trigonometric substitution. However, the given quadratic has no real roots (as the discriminant \( 16 - 36 = -20 \) is negative), which hints that completing the square might simplify the integration.

Complete the square

Rewrite the quadratic expression in the denominator by completing the square. \[ x^2 - 4x + 9 = (x-2)^2 + 5. \] This transforms the expression under the integral into \( \frac{1}{(x-2)^2 + 5} \).

Simplify to a standard integral form

Recognize that the integral \( \int \frac{1}{(x-2)^2 + 5} \, dx \) is similar to the standard form \( \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \arctan\left( \frac{u}{a} \right) + C \), with \( u = x - 2 \) and \( a = \sqrt{5} \).

Change of variables

Let \( u = x - 2 \), which implies \( du = dx \). The integral becomes \( \int \frac{1}{u^2 + (\sqrt{5})^2} \, du \).

Integrate using the arctangent formula

Apply the arctan formula, giving \[ \int \frac{1}{u^2 + (\sqrt{5})^2} \, du = \frac{1}{\sqrt{5}} \arctan\left( \frac{u}{\sqrt{5}} \right) + C. \] Substitute back \( u = x - 2 \) to obtain the solution \[ \frac{1}{\sqrt{5}} \arctan\left( \frac{x-2}{\sqrt{5}} \right) + C. \]

Key concepts

Completing the Square

Completing the square is a useful technique in algebra and calculus that transforms a quadratic expression into a form that is easier to work with, especially for integration or solving equations. Here, we start with a quadratic expression in the denominator:

• The expression is given as \(x^2 - 4x + 9\).
• To complete the square, you take the coefficient of \(x\), which is \(-4\), divide it by 2 to get \(-2\), and then square it to obtain \(4\).
• The expression becomes \((x-2)^2 + (9-4) = (x-2)^2 + 5\).

This rewritten form, \( (x-2)^2 + 5 \), is structured so it resembles part of a standard form useful for integration. Completing the square simplifies the integral and reveals opportunities for substitution.

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integration by replacing variables with trigonometric functions. It often helps with integrals involving quadratic expressions. In our scenario, after completing the square, we have \(\frac{1}{(x-2)^2 + 5}\). Notice:

• This expression resembles the form \( \frac{1}{u^2 + a^2} \).
• The standard trigonometric substitution involves using tangent and arctangent functions.

By recognizing this, we see that a substitution can convert the integral into a form readily integrable with the help of trigonometric identities. Here, we set \(u = x - 2\), simplifying our work with the modified limits of integration and easing the function into a more manageable form.

Indefinite Integral

An indefinite integral represents a family of functions and is expressed without upper and lower limits of integration. In calculus, it is indicated by the integral sign followed by a function and the differential of a variable, as seen in \[ \int \frac{1}{(x-2)^2 + 5} \, dx \].

• The solution of an indefinite integral contains an arbitrary constant \( C \), representing a general form of all possible antiderivatives.
• In this problem, we aim to find an antiderivative that, when derived, returns the original function \( \frac{1}{(x-2)^2 + 5} \).

The process transforms the varying expression into the solution: \( \frac{1}{\sqrt{5}} \arctan\left( \frac{x-2}{\sqrt{5}} \right) + C \), showing the indefinite nature through the constant portion.

Arctan Formula

The arctan formula is crucial in integrals involving expressions of the form \( \frac{1}{a^2 + u^2} \). It connects the integration of such expressions to inverse trigonometric functions, specifically the arctangent. For our expression \[ \int \frac{1}{u^2 + (\sqrt{5})^2} \, du \], the arctan formula applies:

• This standard result says integral is \( \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \).
• Here, \(a\) is \(\sqrt{5}\).

Applying the formulation transforms this integral into a solvable expression using known techniques. The use of the arctangent function simplifies the solution, converting it from a complex integral to an expression that can still be easily manipulated or applied as necessary.