Definite integral 1/(x²−4x+13) from 2 to 5

Q: What to do if the denominator has no real roots?

You need to complete the square: x² + px + q = (x + p/2)² + (q − p²/4). If q − p²/4 > 0, we get a sum of a square and a positive number, and the integral reduces to the tabular ∫ du/(u² + a²) = (1/a)·arctg(u/a) + C.

This problem asks to evaluate the definite integral of a rational function whose denominator has no real roots.

Problem

$\int_{2}^{5} \frac{1}{x^{2} - 4 x + 13} d x$

Solution

Step 1

Complete the square in the denominator:

Why do we complete the square?

The quadratic x² − 4x + 13 has a negative discriminant: D = 16 − 52 = −36 < 0. Hence, it has no real roots and cannot be factored into linear factors. In this case, the only way to integrate is to complete the square and reduce the integral to the tabular ∫ du/(u² + a²) = (1/a)·arctg(u/a) + C.

x^{2} - 4 x + 13 = {(x - 2)}^{2} + 9 = {(x - 2)}^{2} + 3^{2}

Step 2

Make the substitution u = x − 2. Then du = d(x − 2) = dx, and the integral reduces to a tabular form:

How do we reduce to a tabular integral?

The tabular integral has the form ∫ du/(u² + a²) = (1/a)·arctg(u/a) + C. After completing the square, our denominator has the form u² + 3², where u = x − 2, a = 3. So we make the substitution u = x − 2 and obtain a tabular integral.

\int_{2}^{5} \frac{d (x - 2)}{{(x - 2)}^{2} + 3^{2}}

Step 3

Apply the tabular formula:

\frac{1}{3} [_{2}^{5} arctg \frac{x - 2}{3}

Step 4

Substitute the limits of integration. Note that arctg 1 = π/4, arctg 0 = 0:

\frac{1}{3} (arctg 1 - arctg 0) = \frac{1}{3} \cdot \frac{π}{4} = \frac{π}{12}

Answer

$\frac{π}{12}$

Frequently asked questions

What to do if the denominator has no real roots?

You need to complete the square: x² + px + q = (x + p/2)² + (q − p²/4). If q − p²/4 > 0, we get a sum of a square and a positive number, and the integral reduces to the tabular ∫ du/(u² + a²) = (1/a)·arctg(u/a) + C.

Why does an arctangent appear?

Because the derivative of the arctangent is 1/(1 + x²). After the substitution u/a = t, we obtain an integral of the form ∫ dt/(1 + t²) = arctg t + C. This is a standard tabular integral.

How do we complete the square in x²−4x+13?

Take the coefficient of x (−4), divide by 2 to get −2. The square of this number is 4. Then x² − 4x + 13 = (x² − 4x + 4) − 4 + 13 = (x − 2)² + 9. Check: expanding (x − 2)² + 9 gives x² − 4x + 4 + 9 = x² − 4x + 13.

Why does the answer contain π?

Because arctg 1 = π/4 and arctg 0 = 0. After substituting the limits of integration, we get (1/3)·(π/4 − 0) = π/12. This is a typical result for integrals with an arctangent whose upper limit gives arctg 1.