Mathematics - Definite Integration Question with Solution | TestHub

Clearly ${\mathrm{I}}_1={\mathrm{I}}_2$ by using $\mathrm{x}=\frac{1}{\mathrm{u}}$ in ${\mathrm{I}}_2$ See, ${\mathrm{I}}_2={\int }_1^{1/\mathrm{x}}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}$ Let $t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$ $I_2={\int }_1^{\mathrm{u}}\frac{-\frac{1}{{\mathrm{u}}^2}\mathrm{du}}{1+\frac{1}{{\mathrm{u}}^2}}={\int }_{\mathrm{u}}^1\frac{\mathrm{du}}{{\mathrm{u}}^2+1}=I_1=\frac{\pi }{4}-{\tan }^{-1}\mathrm{x}$ \end{document}