The value of ∫ 1/x(e)(e((logex)^2+1))^(-1)/e((logex)^2+1)^(-1)+e^((6-logex)^2+1)^(-1)dx e ∈ to (e^2, e^4) is

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

(1) \(\log _{e} 2\)

(2) 2

(3) 1

(4) \(\mathrm{e}^{2}\)

Answer 1

Correct option is (3) 1 

Let \(\ln \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\)

\(I=\int_{2}^{4} \frac{e^{\frac{1}{1+t^{2}}}}{e^{\frac{1}{1+t^{2}}}+e^{\frac{1}{1+(6-t)^{2}}}} d t\)

\(I=\int_{2}^{4} \frac{e^{\frac{1}{1+(6-t)^{2}}}}{\frac{1}{e^{1+(6-t)^{2}}}+e^{\frac{1}{1+t^{2}}}} d t\)

\(2 \mathrm{I}=\int_{2}^{4} \mathrm{dt}=(\mathrm{t})_{2}^{4}=4-2=2\)

\(\mathrm{I}=1\)