Question

If \(f(x) = \int \frac{x}{(x^2+4)(x^2+9)}\,dx\) and \(f(0) = \frac{1}{5}log(\frac{2}{3})\), then \(f(1) =\)

• A. \(\frac{1}{5}log(2)\)
• B. \(\frac{1}{10}log(2)\)
• C. \(-\frac{1}{5}log(2)\)
• D. \(-\frac{1}{10}log(2)\)

Hint:

Substitute u = x squared and split into partial fractions.

Answer 1

The Correct Option is D

Step 1: Different split:
Write $\frac{x}{(x^2+4)(x^2+9)} = \frac{x}{5}\left(\frac{1}{x^2+4} - \frac{1}{x^2+9}\right)$, since $\frac{1}{x^2+4} - \frac{1}{x^2+9} = \frac{5}{(x^2+4)(x^2+9)}$.

Step 2: Integrate term by term:
$\int\frac{x}{x^2+4}dx = \frac12\log(x^2+4)$ and similarly for 9. So $f(x) = \frac{1}{10}[\log(x^2+4) - \log(x^2+9)] + C$.

Step 3: Use f(0):
$\frac1{10}\log\frac49 = \frac15\log\frac23$, so $C = 0$. Then $f(1) = \frac1{10}\log\frac12 = -\frac{1}{10}\log2$.

Final Answer:
f(1) = -(1/10) log 2, option (D). \[ \boxed{-\frac{1}{10}\log 2} \]