The value of _ e^2 ^ e ^4 1 x ( e^ ( ( _ e x )^2+1 )^ -1 e^ ( ( _… - Vidyadip

MCQ

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
$\log _{ c } 2$
B
2
✓
1
D
$e ^2$

✓

Answer

Correct option: C.

1

(C)
Let $\ln x = t \Rightarrow \frac{ dx }{ x }= dt$
$
\begin{array}{l}
I=\int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t)^2}}} d t \\
I=\int_2^4 \frac{\frac{1}{e^{1+(6-t)^2}}}{\frac{1}{e^{1+(6-t)^2}}+e^{\frac{1}{1+t^2}}} d t \\
2 I=\int_2^4 d t=(t)_2^4=4-2=2 \\
I=1
\end{array}
$

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