00:01
Okay, this i want to integrate, and if you have just single log functions like this, you must use parts integration.
00:09
So the integral then of lnx plus 3 times 1, dx.
00:17
Then this here is your u, and this is your dv by dx in the parts formula that you will find on all formula sheets.
00:29
So what we have then is u times v is the first part of the formula.
00:35
Or u is l n of x plus 3, no change there at all.
00:41
Define v, i integrate 1.
00:45
That simply is x.
00:47
I'll put that in brackets here just to be clear.
00:53
Minus then the integral of v, i've just found it, x times du by dx.
01:01
I differentiate this, what i will get is 1 over x plus 3 times 1 by the chamber of d x.
01:13
So what i have then is x ln x plus 3 minus the integral of x over x plus 3 d x.
01:28
Now, this part here i will do by substitution.
01:36
I will let u equal x plus 3, which means du by dx will be 1, which means then that du is the same value as dx.
01:52
So what i have here then is the integral of x will be u minus 3 from here.
02:02
Let's rearrange it over u and dx equals to u.
02:10
So that becomes the integral of 1 minus 3 over u to u...