JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 15

Pair up. With $t\to\frac1t$, $f\!\left(\frac1\alpha\right)=\displaystyle\int_1^\alpha\frac{\log_{10}t}{t(1+t)}dt$, so $f(\alpha)+f\!\left(\frac1\alpha\right)=\int_1^\alpha\frac{\log_{10}t}t\,dt=\frac{(\ln\alpha)^2}{2\ln10}$.

At $e^3$: $\dfrac9{2\log_e10}$.

(D)