JEE Main 2026 — 5 April, Morning Shift — Question 41

Put $\mathrm{x}=2 \mathrm{t} \Rightarrow \mathrm{dx}=2 \mathrm{dt}$ $\mathrm{I}=\int_{0}^{\infty} \frac{\ell \mathrm{n} 2 \mathrm{t}}{4 \mathrm{t}^{2}+4}(2 \mathrm{dt})=\frac{1}{2} \int_{0}^{\infty} \frac{\ell \mathrm{n} 2+\ell \mathrm{nt}}{\mathrm{t}^{2}+1} \mathrm{dt}$ $=\frac{1}{2} \int_{0}^{\infty} \frac{\ell \mathrm{n} 2}{\mathrm{t}^{2}+1} \mathrm{dt}+\frac{1}{2} \int_{0}^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^{2}+1} \mathrm{dt}$ $\left[\frac{\ln 2}{2} \tan ^{-1} \mathrm{t}\right]_{0}^{\infty}+\mathrm{I}_{1}$ $=\frac{\ell \mathrm{n} 2}{2} \cdot \frac{\pi}{2}+\mathrm{I}_{1}$ $\mathrm{I}_{2}=\frac{1}{2} \int_{0}^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^{2}+1} \mathrm{dt} \mathrm{t}=\frac{1}{\mathrm{u}} \Rightarrow \mathrm{dt}=-\frac{\mathrm{du}}{\mathrm{u}^{2}}$ $=\frac{1}{2} \int_{\infty}^{0} \frac{\ell \mathrm{n}(1 / \mathrm{u})}{\frac{1}{\mathrm{u}^{2}}+1}\left(-\frac{\mathrm{du}}{\mathrm{u}^{2}}\right)$ Add $\Rightarrow \mathrm{I}_{1}=0$ $\mathrm{I}=\frac{\pi \ell \mathrm{n} 2}{4}$