Question

Let f(x) = \(\int \frac{x}{(x^2+1)(x^2+3)} dx\). If f(3) = \(\frac{1}{4} \log(\frac{5}{6})\), then f(0) =

• A. \(\frac{1}{4}\log\left(\frac{1}{3}\right)\)
• B. \(0\)
• C. \(\frac{1}{2}\log\left(\frac{1}{3}\right)\)
• D. \(\log\left(\frac{1}{3}\right)\)

Hint:

Use substitution and partial fractions to evaluate the integral.

Answer 1

The Correct Option is A

Step 1: Use substitution to simplify the integral.
Let u = x^2. Then du = 2x dx, so x dx = \(\frac{1}{2}\) du.
f(x) = \(\int \frac{x}{(x^2+1)(x^2+3)} dx = \int \frac{1}{2(u+1)(u+3)} du\)

Step 2: Use partial fractions.
\(\frac{1}{(u+1)(u+3)} = \frac{A}{u+1} + \frac{B}{u+3}\)
\(1 = A(u+3) + B(u+1)\)
When u = -1, \(1 = 2A \Rightarrow A = \frac{1}{2}\).
When u = -3, \(1 = -2B \Rightarrow B = -\frac{1}{2}\).
\(\frac{1}{(u+1)(u+3)} = \frac{1}{2(u+1)} - \frac{1}{2(u+3)}\)

Step 3: Integrate with respect to u.
f(x) = \(\frac{1}{2} \int \left(\frac{1}{2(u+1)} - \frac{1}{2(u+3)}\right) du\)
f(x) = \(\frac{1}{4} \int \left(\frac{1}{u+1} - \frac{1}{u+3}\right) du\)
f(x) = \(\frac{1}{4} (\log|u+1| - \log|u+3|) + C\)
f(x) = \(\frac{1}{4} \log\left|\frac{u+1}{u+3}\right| + C\)

Step 4: Substitute back u = x^2.
f(x) = \(\frac{1}{4} \log\left|\frac{x^2+1}{x^2+3}\right| + C\)

Step 5: Use the given condition f(3) = \(\frac{1}{4} \log(\frac{5}{6})\).
f(3) = \(\frac{1}{4} \log\left|\frac{3^2+1}{3^2+3}\right| + C = \frac{1}{4} \log\left(\frac{10}{12}\right) + C = \frac{1}{4} \log\left(\frac{5}{6}\right) + C\)
\(\frac{1}{4} \log\left(\frac{5}{6}\right) = \frac{1}{4} \log\left(\frac{5}{6}\right) + C\)
C = 0

Step 6: Find f(0).
f(x) = \(\frac{1}{4} \log\left|\frac{x^2+1}{x^2+3}\right|\)
f(0) = \(\frac{1}{4} \log\left|\frac{0^2+1}{0^2+3}\right| = \frac{1}{4} \log\left(\frac{1}{3}\right)\)
Therefore, f(0) = \(\frac{1}{4}\log\left(\frac{1}{3}\right)\).