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Definite Integration
JEE Main 2025
23 January
Morning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration
JEE Main 2025 · 23 January · Morning Shift · Mathematics · Definite Integration
The value of ∫_ e^2 ^ e^4 (1)/(x)( e^ (( _e x)^2+1)^ -1 e^ (( _e...
The value of
∫
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2
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2
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6
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\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
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Options
A
2
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2
B
log
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\log _e 2
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C
1
1
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D
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Correct answer
C.
1
1
1
Step-by-step solution
Put
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…
(i)
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(ii)
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Using
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\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 \frac{e^{\left(\mathrm{t}^2+1\right)^{-1}}}{e^{\left(\mathrm{t}^2+1\right)^{-1}}+e^{\left((6-t)^2+1\right)^{-1}}} d t \quad \ldots \text{ (i) } \\ & I=\int_2^4 \frac{e^{\left((6-t)^2+1\right)^{-1}}}{e^{\left((6-t)^2+1\right)^{-1}}+e^{\left(t^2+1\right)^{-1}} d t} \ldots \text{ (ii) } \\ & \left\{\text{ Using } \int_a^0 f(x) d x=\int_a^0 f(a+b-x) d x\right\} \end{aligned}
Put
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…
(i)
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(ii)
{
Using
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Adding (i) and (ii) gives
2
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2 l=\int d t \Rightarrow l=1
2
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∫
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⇒
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1
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