00:01
Problem 35 says that we have to integrate lnx over square dx.
00:12
So let's substitute lnx as t.
00:15
So this means that dx over x is equal to dt and this also means that the value of x is a less to the power t.
00:28
So if we'll use all these, the integral i now becomes replacing lnx by t.
00:36
And in fact let's rewrite our integral first to make it more simpler so integral is lnx d x and x square can be splited as x times x now this is d x over x first lnx is replaced by t d x over x is replaced by d t and then we have this x which is e raised to the power t so integral becomes t e a raise to the power minus t d t integral now, using integration by parts, this will be the first term, this will be the second term.
01:10
So the integral will be the first term as it is, integration of second term will be minus e days to the power minus t.
01:16
Minus integral of differentiation of t is one.
01:19
Integration of e to the power minus t is minus e days to the power minus t and this one is again integrated...