Question Number 200747 by Rupesh123 last updated on 22/Nov/23
Answered by MM42 last updated on 22/Nov/23
![lnx=u⇒∫_1 ^∞ (du/u^2 ) =−(1/u)]_1 ^∞ =1 ✓](https://www.tinkutara.com/question/Q200755.png)
$$\left.{lnx}={u}\Rightarrow\int_{\mathrm{1}} ^{\infty} \:\frac{{du}}{{u}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{{u}}\right]_{\mathrm{1}} ^{\infty} =\mathrm{1}\:\checkmark \\ $$
Commented by esmaeil last updated on 23/Nov/23
![=−(1/(lnx))=(1/(−lnx))=(1/(ln(1/x)))]_e ^∞ =^(?????) 1](https://www.tinkutara.com/question/Q200777.png)
$$\left.=−\frac{\mathrm{1}}{{lnx}}=\frac{\mathrm{1}}{−{lnx}}=\frac{\mathrm{1}}{{ln}\frac{\mathrm{1}}{{x}}}\underset{{e}} {\overset{\infty} {\right]}}\overset{?????} {=}\mathrm{1} \\ $$
Commented by MM42 last updated on 23/Nov/23
![or −(1/(lnx))]_e ^∞ =−(1/(ln∞))+(1/(lne))=1](https://www.tinkutara.com/question/Q200783.png)
$${or} \\ $$$$\left.−\frac{\mathrm{1}}{{lnx}}\right]_{{e}} ^{\infty} =−\frac{\mathrm{1}}{{ln}\infty}+\frac{\mathrm{1}}{{lne}}=\mathrm{1} \\ $$
Commented by MM42 last updated on 23/Nov/23
![lnx=u⇒if x→e⇒u→1 & x→∞⇒u→∞ ⇒−(1/u)]_1 ^∞ =−(1/∞)+(1/1)=1](https://www.tinkutara.com/question/Q200782.png)
$${lnx}={u}\Rightarrow{if}\:\:{x}\rightarrow{e}\Rightarrow{u}\rightarrow\mathrm{1}\:\:\&\:\:{x}\rightarrow\infty\Rightarrow{u}\rightarrow\infty \\ $$$$\left.\Rightarrow−\frac{\mathrm{1}}{{u}}\right]_{\mathrm{1}} ^{\infty} \:=−\frac{\mathrm{1}}{\infty}+\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1}\: \\ $$
Answered by Calculusboy last updated on 28/Nov/23
![Solution: let u=Inx xdu=dx when x=∞ u=∞ and when x=e u=1 I=∫_1 ^∞ ((xdu)/(x(u)^2 )) ⇔ I=∫_1 ^∞ (du/u^2 ) I=−[(1/u)]_1 ^∞ +C I=−[(1/∞)−(1/1)] I=−[0−1] I=1 ∴∫_e ^∞ (1/(x(In(x))^2 ))dx=1](https://www.tinkutara.com/question/Q201013.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{Inx}}\:\:\:\:\boldsymbol{{xdu}}=\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\infty\:\boldsymbol{{u}}=\infty\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\boldsymbol{{e}}\:\:\boldsymbol{{u}}=\mathrm{1} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{{xdu}}}{\boldsymbol{{x}}\left(\boldsymbol{{u}}\right)^{\mathrm{2}} }\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{1}} ^{\infty} \frac{\boldsymbol{{du}}}{\boldsymbol{{u}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{I}}=−\left[\frac{\mathrm{1}}{\boldsymbol{{u}}}\right]_{\mathrm{1}} ^{\infty} +\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=−\left[\frac{\mathrm{1}}{\infty}−\frac{\mathrm{1}}{\mathrm{1}}\right] \\ $$$$\boldsymbol{{I}}=−\left[\mathrm{0}−\mathrm{1}\right] \\ $$$$\boldsymbol{\mathrm{I}}=\mathrm{1} \\ $$$$\therefore\int_{\boldsymbol{{e}}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}\left(\boldsymbol{{In}}\left(\boldsymbol{{x}}\right)\right)^{\mathrm{2}} }\boldsymbol{{dx}}=\mathrm{1} \\ $$$$ \\ $$