Question Number 137907 by peter frank last updated on 08/Apr/21
Answered by EnterUsername last updated on 08/Apr/21
![∫_1 ^e ((1+lnx)/x)dx=∫_1 ^e ((1/x)+((lnx)/x))dx =[lnx+((ln^2 x)/2)]_1 ^e =1+(1/2)=(3/2) (Notice ((dlnx)/dx)=(1/x))](https://www.tinkutara.com/question/Q137909.png)
$$\int_{\mathrm{1}} ^{{e}} \frac{\mathrm{1}+{lnx}}{{x}}{dx}=\int_{\mathrm{1}} ^{{e}} \left(\frac{\mathrm{1}}{{x}}+\frac{{lnx}}{{x}}\right){dx} \\ $$$$=\left[{lnx}+\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{2}}\right]_{\mathrm{1}} ^{{e}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left({Notice}\:\frac{{dlnx}}{{dx}}=\frac{\mathrm{1}}{{x}}\right) \\ $$
Commented by peter frank last updated on 08/Apr/21
$${thank}\:{you} \\ $$