Q. The value of $\int_{e^2}^{e^4}\frac1x\left(\frac{e^{\left((\log_ex)^2+1\right)^{-1}}}{e^{\left((\log_ex)^2+1\right)^{-1}}+e^{\left((6-\log_ex)^2+1\right)^{-1}}}\right)dx$ is
A) 2
B) $\log_e2$
C) 1
D) $e^2$
Explanation
[Solution by Dr. Abubacker] Put $t=\ln x$: $I=\int_2^4\frac{g(t)}{g(t)+g(6-t)}dt$ with $g(t)=e^{1/(t^2+1)}$. Replacing $t\to6-t$ (limits $2\leftrightarrow4$) gives the complementary integrand, so $2I=\int_2^41\,dt=2$ and $I=1$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.