Question Number 73030 by Tanmay chaudhury last updated on 05/Nov/19
Answered by Tanmay chaudhury last updated on 05/Nov/19
$$\int_{\mathrm{0}} ^{\mathrm{4}} \frac{{ln}\mathrm{2}}{{lnx}}−\frac{{ln}\mathrm{2}×{ln}\mathrm{2}}{{lnx}×{lnx}×{ln}\mathrm{2}}{dx} \\ $$$${ln}\mathrm{2}\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\mathrm{1}}{{lnx}}−\frac{\mathrm{1}}{\left({lnx}\right)^{\mathrm{2}} }{dx} \\ $$$${ln}\mathrm{2}\int_{\mathrm{2}} ^{\mathrm{4}} \frac{{lnx}−\mathrm{1}}{\left({lnx}\right)^{\mathrm{2}} }{dx} \\ $$$${ln}\mathrm{2}\int_{\mathrm{2}} ^{\mathrm{4}} \frac{{lnx}.\frac{{dx}}{{dx}}−{x}.\frac{{d}}{{dx}}\left({lnx}\right)\:\:}{\left({lnx}\right)^{\mathrm{2}} }{d}\overset{} {{x}} \\ $$$${ln}\mathrm{2}\int_{\mathrm{2}} ^{\mathrm{4}} \frac{{d}}{{dx}}\left(\frac{{x}}{{lnx}}\right){dx}\: \\ $$$${ln}\mathrm{2}×\mid\frac{{x}}{{lnx}}\mid_{\mathrm{2}} ^{\mathrm{4}} \\ $$$${ln}\mathrm{2}×\left(\frac{\mathrm{4}}{{ln}\mathrm{4}}−\frac{\mathrm{2}}{{ln}\mathrm{2}}\right) \\ $$$${ln}\mathrm{2}×\left(\frac{\mathrm{4}}{\mathrm{2}{ln}\mathrm{2}}−\frac{\mathrm{2}}{{ln}\mathrm{2}}\right)=\mathrm{0}\:\:{answer} \\ $$
Answered by mind is power last updated on 05/Nov/19
![∫_2 ^4 (log_x (2)−(((log_x (2))^2 )/(ln(2))))dx =∫((ln(2))/(ln(x)))−((ln(2))/(ln^2 (x)))dx =ln(2)∫((ln(x)−1)/(ln^2 (x)))dx =ln(2)∫d((x/(ln(x))))=ln2[(x/(ln(x)))]_2 ^4 =((4ln(2))/(ln(4)))−((2ln(2))/(ln(2)))=2−2=0](https://www.tinkutara.com/question/Q73060.png)
$$\int_{\mathrm{2}} ^{\mathrm{4}} \left(\mathrm{log}_{\mathrm{x}} \left(\mathrm{2}\right)−\frac{\left(\mathrm{log}_{\mathrm{x}} \left(\mathrm{2}\right)\right)^{\mathrm{2}} }{\mathrm{ln}\left(\mathrm{2}\right)}\right)\mathrm{dx} \\ $$$$=\int\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{x}\right)}−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)\int\frac{\mathrm{ln}\left(\mathrm{x}\right)−\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)\int\mathrm{d}\left(\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{x}\right)}\right)=\mathrm{ln2}\left[\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{x}\right)}\right]_{\mathrm{2}} ^{\mathrm{4}} =\frac{\mathrm{4ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{4}\right)}−\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}=\mathrm{2}−\mathrm{2}=\mathrm{0} \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
$${thank}\:{you}\:{sir}… \\ $$
Commented by mind is power last updated on 05/Nov/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$$$ \\ $$