Problem 57 - Calculus for Biology and Medicine

Question

Complete the square in the denominator and evaluate the integral. $$ \int \frac{1}{x^{2}-4 x+13} d x $$

Step-by-Step Solution

Answer

\(\frac{1}{3} \arctan\left(\frac{x-2}{3}\right) + C\)

Step 1: Identify the Denominator as a Quadratic

The expression in the denominator, \(x^2 - 4x + 13\), is a quadratic polynomial. This is the part of the integrand we want to simplify using completing the square.

Step 2: Completing the Square

To complete the square for \(x^2 - 4x + 13\), rewrite \(x^2 - 4x\). Take half of the linear coefficient \(-4\), square it, and add and subtract that value inside the quadratic. Half of \(-4\) is \(-2\) and \((-2)^2 = 4\). Thus, rewrite: \(x^2 - 4x + 13 = (x^2 - 4x + 4) + 13 - 4.\) This simplifies to \((x-2)^2 + 9.\)

Step 3: Rewrite the Integral

Now that we have \((x-2)^2 + 9\) in the denominator, rewrite the integral as: \[\int \frac{1}{(x-2)^2 + 9} \, dx.\] This is now in a form that resembles the standard integral for arctangent.

Step 4: Use the Arctangent Integral Formula

Recognize that \(\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C\) is the standard form. Here, \(u = x-2\) and \(a^2 = 9\), so \(a = 3\). Substitute: \[x-2=u \,\Rightarrow\, du = dx.\] Thus, the integral becomes: \[\int \frac{1}{u^2 + 3^2} \, du = \frac{1}{3} \arctan\left(\frac{u}{3}\right) + C.\]

Step 5: Substitute Back the Original Variable

Since \(u = x-2\), substitute back to get the antiderivative in terms of \(x\): \[\frac{1}{3} \arctan\left(\frac{x-2}{3}\right) + C.\]

Step 6: Conclusion

The integral of the given function after completing the square and evaluating using the arctangent formula is: \[\int \frac{1}{x^2 - 4x + 13} \, dx = \frac{1}{3} \arctan\left(\frac{x-2}{3}\right) + C.\]

Key Concepts

Completing the Square

When faced with a quadratic polynomial, like in the expression \(x^2 - 4x + 13\), one handy tool is completing the square. This technique helps simplify the expression into a form that's easier to work with, especially when dealing with integrals. The idea is to transform the polynomial into a perfect square trinomial plus some constant. Here's how you can do it:

• Look at the quadratic and linear terms, \(x^2 - 4x\).
• Take half of the linear coefficient (which is \(-4\)), giving you \(-2\).
• Square this half to get \(4\), which you'll add and subtract within the polynomial.

By doing this, \(x^2 - 4x + 13\) becomes \((x-2)^2 + 9\). This new form neatly separates the squared component and makes it much simpler to integrate by reducing it to a recognizable formula.

Arctangent Integral Formula

Once a quadratic polynomial is expressed in perfect square form, like \((x-2)^2 + 9\), it opens the way to use the arctangent integral formula. This standard formula helps you find integrals of the type \(\int \frac{1}{u^2 + a^2} \, du\). It states:
\[\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C\]
Here, \(u\) and \(a\) are parts of the perfect square form. For instance, in \((x-2)^2 + 9\), \(u = x-2\) and \(a^2 = 9\), so \(a = 3\). Substituting these values into the formula allows us to compute the integral by recognizing it in the form required for the arctangent result.

Quadratic Polynomial Integration

Integrating a quadratic polynomial involves converting it into a manageable form like a perfect square. Initially, a polynomial such as \(x^2 - 4x + 13\) may not seem friendly for direct integration. Using the process of completing the square, we converted it into \((x-2)^2 + 9\). This form is easier to integrate because it fits the arctangent integral formula described earlier.
Once the square form is achieved and the integral evaluated using the formula, we backtrack to substitute the original variable, making sure the final antiderivative is in terms of \(x\).
This technique effectively makes complex integrations more approachable by using simple algebraic manipulations to qualify them for standard integration formulas.