JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of
\(\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x\) is
- A \(2\)
- B \(\log _e 2\)
- C \(1\)
- D \(e^2\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {Put } \ln x=t \Rightarrow \frac{1}{x} d x=d t \quad \begin{array}{|c|c|} \hline x & t \\ \hline e^2 & 2 \\ \hline e^4 & 4 \\ \hline \end{array} \\ & I=\int_2^4…
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