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

Answer: (D)

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

Answer 2

The correct option is (D) : $\frac{1}{2}\left(\log _e 17-\log _e 19\right)$
Put x2=t
∫$\frac{dt}{(t+1)(t+3)}$
​=$\frac{1}{2}$​∫($\frac{1}{t+1}$​−$\frac{1}{t+3}$​)dt
f(x)=$\frac{1}{2}$​ln($\frac{x^2+1}{x^2+3}$)+C
f(3)=$\frac{1}{2}$​(ln10−ln12)+C
⇒C=0
f(4)=$\frac{1}{2}$​ln($\frac{17}{19}$​)