The value of $\int_{e^{2}}^{e^{4}}\frac{1}{x}\left(\frac{e^{\frac{1}{1+(\ln x)^{2}}}}{e^{\frac{1}{1+(\ln x)^{2}}}+e^{\frac{1}{1+(6-\ln x)^{2}}}}\right)dx$ is
(1) $\log_{e}2$ (2) 2 (3) 1 (4) $e^{2}$
Solution:
Let $I=\int_{e^{2}}^{e^{4}}\frac{1}{x}\left(\frac{e^{\frac{1}{1+(\ln x)^{2}}}}{e^{\frac{1}{1+(\ln x)^{2}}}+e^{\frac{1}{1+(6-\ln x)^{2}}}}\right)dx$
Put $\ln x=t \implies \frac{1}{x}dx=dt$.
Limits: $x=e^2 \to t=2$, $x=e^4 \to t=4$.
$$I=\int_{2}^{4}\frac{e^{\frac{1}{1+t^{2}}}}{e^{\frac{1}{1+t^{2}}}+e^{\frac{1}{1+(6-t)^{2}}}}dt \quad \dots(1)$$
Using property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ where $a+b=6$:
$$I=\int_{2}^{4}\frac{e^{\frac{1}{1+(6-t)^{2}}}}{e^{\frac{1}{1+(6-t)^{2}}}+e^{\frac{1}{1+t^{2}}}}dt \quad \dots(2)$$
Adding (1) and (2):
$$2I=\int_{2}^{4}1 \cdot dt = [t]_{2}^{4} = 4-2=2$$
$$I=1$$
Ans. (3)