JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 7)
Let $$f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx} $$. If $$f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)$$, then $$f(4)$$ is equal to
$${\log _e}19 - {\log _e}20$$
$${\log _e}17 - {\log _e}18$$
$${1 \over 2}({\log _e}19 - {\log _e}17)$$
$${1 \over 2}({\log _e}17 - {\log _e}19)$$
Explanation
$f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$
$$ \begin{aligned} & \text { Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\ & f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\ & =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right)+C \Rightarrow \\\\ & f(3)=\frac{1}{2} \log _{e}\left(\frac{10}{12}\right)+C \\\\ & \because f(3)+\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right) \Rightarrow C=0 \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right) \Rightarrow \\\\ & f(4)=\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right) \end{aligned} $$
$$ \begin{aligned} & \text { Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\ & f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\ & =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right)+C \Rightarrow \\\\ & f(3)=\frac{1}{2} \log _{e}\left(\frac{10}{12}\right)+C \\\\ & \because f(3)+\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right) \Rightarrow C=0 \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right) \Rightarrow \\\\ & f(4)=\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right) \end{aligned} $$
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