Question Number 83077 by mhmd last updated on 27/Feb/20
$$\mathrm{If}\:\:\:{I}_{\mathrm{1}} =\underset{{e}} {\overset{{e}^{\mathrm{2}} } {\int}}\:\frac{{dx}}{\mathrm{log}\:{x}}\:\:\mathrm{and}\:\:{I}_{\mathrm{2}} =\:\underset{\:\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{{e}^{{x}} }{{x}}\:{dx},\:\mathrm{then} \\ $$
Answered by TANMAY PANACEA last updated on 28/Feb/20
$${t}={lnx}\:\:\:\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{x}}\rightarrow{e}^{{t}} {dt}={dx}\:{so}\:\boldsymbol{{I}}_{\mathrm{1}} =\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{e}^{{t}} {dt}}{{t}}={I}_{\mathrm{2}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} ={I}_{\mathrm{2}} \\ $$$${now}\:\:{e}^{{t}} =\mathrm{1}+{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}+{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}+…}{{t}}{dt} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{{t}}+\mathrm{1}+\frac{{t}}{\mathrm{2}!}+\frac{{t}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{4}!}+…\right){dt} \\ $$$$\mid\frac{{lnt}}{\mathrm{1}}+\frac{{t}}{\mathrm{1}}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!×\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!×\mathrm{3}}+…\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$={ln}\left(\frac{\mathrm{2}}{\mathrm{1}}\right)+\left(\mathrm{2}−\mathrm{1}\right)+\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}!×\mathrm{2}}+\frac{\mathrm{2}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\mathrm{3}!×\mathrm{3}}+… \\ $$