Question

Let $f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$. If $f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)$, then $f(4)$ is equal to

• A. $\log _{ e } 17-\log _{ e } 18$
• B. $\log _e 19-\log _e 20$
• C. $\frac{1}{2}\left(\log _e 19-\log _e 17\right)$
• D. $\frac{1}{2}\left(\log _e 17-\log _e 19\right)$

Answer 1

The Correct Option is D

To solve the given problem, we start by evaluating the integral function:

f(x)=\int \frac{2x}{(x^2+1)(x^2+3)} \, dx

The integral can be simplified using partial fraction decomposition. Setting:

\frac{2x}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}

We equate coefficients upon expanding and equating to determine A, B, C, and D. Our goal is to solve for these constants and then integrate each term separately. Finally, we calculate:

Given: f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)

If the setup of partial fractions is done correctly, any intermediate simplification will reveal:

f(x) = \frac{1}{2} \left(\log _e (x^2+1) - \log _e (x^2+3) \right) + C

Given the condition, substitute x = 3:

f(3) = \frac{1}{2}(\log_e (3^2 + 1) - \log_e (3^2 + 3)) = \frac{1}{2}(\log_e 10 - \log_e 12)

This simplifies to match the given:

\frac{1}{2}\left(\log _e 5-\log _e 6\right)

Now we evaluate f(4):

f(4) = \frac{1}{2}(\log_e (4^2 + 1) - \log_e (4^2 + 3))

Then calculate:

f(4) = \frac{1}{2}(\log_e 17 - \log_e 19)

Hence the correct answer is:

$\frac{1}{2}\left(\log _e 17-\log _e 19\right)$