Verified Solution • 2025–26 Session moderate

QJEEM23Ma329a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma329a Solution: Let . If , then is equal to

Let f(x)=∫2x(x2+1)(x2+3)dxf(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx} . If f(3)=12(log⁡e5−log⁡e6)f(3) = {1 \over 2}({\log _e}5 - {\log _e}6) , then f(4)f(4) is equal to

Multiple Choice Options

  • A log⁡e19−log⁡e20{\log _e}19 - {\log _e}20
  • B log⁡e17−log⁡e18{\log _e}17 - {\log _e}18
  • C 12(log⁡e19−log⁡e17){1 \over 2}({\log _e}19 - {\log _e}17)
  • D 12(log⁡e17−log⁡e19){1 \over 2}({\log _e}17 - {\log _e}19)

Step-by-Step Verified Solution

f(x)=∫2x(x2+1)(x2+3)dxf(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x

amp; Put x2=t⇒2xdx=dtamp;f(x)=∫dt(t+1)(t+3)=∫dt(t+2)2−1amp;=12log⁡e∣t+1t+3∣+Camp;f(x)=12log⁡e(x2+1x2+3)+C⇒amp;f(3)=12log⁡e(1012)+Camp;∵f(3)+12(log⁡e5−log⁡e6)⇒C=0amp;f(x)=12log⁡e(x2+1x2+3)⇒amp;f(4)=12(log⁡e17−log⁡e19)\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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