The value of _e^2^e^4 (1)/(x)(e^(( _e x)^2+1)^-1e^(( _e x)^2+1)^-1+e^((6- _e x)^2+1)^-1) d…

Question

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

Accepted Answer

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

Solutions · 1