Question

Evaluate: \( \int_1^2 \frac{dx}{x^2+6x+5} \)

Hint:

For rational functions \( \frac{P(x)}{Q(x)} \), if the denominator \( Q(x) \) can be factored into linear terms, partial fraction decomposition is the standard method to simplify the integrand into a sum of simpler fractions.

Answer 1

Step 1: Factor the denominator and set up the partial fraction decomposition. \[ x^2+6x+5 = (x+1)(x+5) \] \[ \frac{1}{(x+1)(x+5)} = \frac{A}{x+1} + \frac{B}{x+5} \] Multiplying by the denominator gives \( 1 = A(x+5) + B(x+1) \).
Step 2: Solve for A and B.
If \( x = -1 \), then \( 1 = A(4) \implies A = \frac{1}{4} \).
If \( x = -5 \), then \( 1 = B(-4) \implies B = -\frac{1}{4} \).
Step 3: Rewrite the integral with the partial fractions. \[ \int_1^2 \left(\frac{1/4}{x+1} - \frac{1/4}{x+5}\right) dx = \frac{1}{4} \int_1^2 \left(\frac{1}{x+1} - \frac{1}{x+5}\right) dx \] Step 4: Integrate and evaluate the definite integral. \[ \frac{1}{4} [\ln|x+1| - \ln|x+5|]_1^2 = \frac{1}{4} \left[\ln\left|\frac{x+1}{x+5}\right|\right]_1^2 \] \[ = \frac{1}{4} \left( \ln\left|\frac{2+1}{2+5}\right| - \ln\left|\frac{1+1}{1+5}\right| \right) = \frac{1}{4} \left( \ln\left(\frac{3}{7}\right) - \ln\left(\frac{2}{6}\right) \right) \] \[ = \frac{1}{4} \left( \ln\left(\frac{3}{7}\right) - \ln\left(\frac{1}{3}\right) \right) = \frac{1}{4} \ln\left(\frac{3/7}{1/3}\right) = \frac{1}{4} \ln\left(\frac{9}{7}\right) \]