Question

If \(\int g(x)dx=f(x)\) then \(\int \frac{g(x)}{9-[f(x)]^{2}}dx\) is equal to

• A. \(\frac{1}{6}\log |\frac{3-g(x)}{3-f(x)}|+c\)
• B. \(\frac{1}{3}\log |\frac{3+f(x)}{3-f(x)}|+c\)
• C. \(\frac{1}{6}\log |\frac{3+f(x)}{3-f(x)}|+c\)
• D. \(\frac{1}{3}\log |\frac{3+f(x)}{3+g(x)}|+c\)

Hint:

Substitute u = f(x) so that g(x) dx = du and use the standard formula.

Answer 1

The Correct Option is C

Step 1: Eliminate by Form
The answer must be written in terms of $f(x)$, because the substitution $u=f(x)$ removes $g(x)$. This removes options (A) and (D).

Step 2: Check the Constant
The standard form with $a=3$ has the constant $\dfrac{1}{2a}=\dfrac{1}{6}$, so option (B) with $\dfrac{1}{3}$ is wrong.

Step 3: Select
Only option (C) has the factor $\dfrac{1}{6}$ and the ratio $\dfrac{3+f}{3-f}$.

Final Answer:
Option (C) is the only form that matches the standard result. \[ \boxed{\text{(C) }\dfrac{1}{6}\log\left|\dfrac{3+f(x)}{3-f(x)}\right|+c} \]