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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 ✓

lnx=u⇒∫1∞duu2=−1u]1∞=1✓

Commented by esmaeil last updated on 23/Nov/23

=−(1/(lnx))=(1/(−lnx))=(1/(ln(1/x)))]_e ^∞ =^(?????) 1

Missing \left or extra \right

Commented by MM42 last updated on 23/Nov/23

or −(1/(lnx))]_e ^∞ =−(1/(ln∞))+(1/(lne))=1

or−1lnx]e∞=−1ln∞+1lne=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

lnx=u⇒ifx→e⇒u→1&x→∞⇒u→∞⇒−1u]1∞=−1∞+11=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

Solution:letu=Inxxdu=dxwhenx=∞u=∞andwhenx=eu=1I=∫1∞xdux(u)2⇔I=∫1∞duu2I=−[1u]1∞+CI=−[1∞−11]I=−[0−1]I=1∴∫e∞1x(In(x))2dx=1

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