Help me with this Definite integral: Int, from 1 to (n+1), of ln[x] dx

[Shaa-Maan]

ok so i was going through some exercises from my book, and i've encountered this beast:

Int from 1 to (n+1) of ln[x]dx

So I've calculated the indefinite integral by parts, and it spit out [xln[x] -x +constant]. I'm confident that this is right, because i've encountered this one numerous times.
then i went the usual way, that is:

int from a to b f(x)dx = F(b)-F(a), so in this case:

[xln[x] -x] from 1 to (n+1) =
={ [ (n+1)*ln(n+1) - (n+1) ] - [ 1*ln(1) - 1 ] } =
={ [ (n+1)*ln(n+1) - (n+1) ] - [ 1*0 - 1] }=
={ [ (n+1)*ln(n+1) - (n+1) ] + 1}=
=(n+1) * [ ln(n+1) - 1 ] + 1

but my book says it's supposed to be ln(n!).

I have a hard time believing that they're equal.

So my question is: Are they really equal, or am I doing something wrong here?

thx for help in advance.

 

[Deleted member 4993]

ok so i was going through some exercises from my book, and i've encountered this beast:

Int from 1 to (n+1) of ln[x]dx

So I've calculated the indefinite integral by parts, and it spit out [xln[x] -x +constant]. I'm confident that this is right, because i've encountered this one numerous times.
then i went the usual way, that is:

int from a to b f(x)dx = F(b)-F(a), so in this case:

[xln[x] -x] from 1 to (n+1) =
={ [ (n+1)*ln(n+1) - (n+1) ] - [ 1*ln(1) - 1 ] } =
={ [ (n+1)*ln(n+1) - (n+1) ] - [ 1*0 - 1] }=
={ [ (n+1)*ln(n+1) - (n+1) ] + 1}=
=(n+1) * [ ln(n+1) - 1 ] + 1

but my book says it's supposed to be ln(n!).

I have a hard time believing that they're equal.

So my question is: Are they really equal, or am I doing something wrong here?

thx for help in advance.

?
Can you expand ln(n!)?

What do you get?

 

[pka]

ok so i was going through some exercises from my book, and i've encountered this beast:
Int from 1 to (n+1) of ln[x]dx
but my book says it's supposed to be ln(n!).

If you agree that it is the case that $\displaystyle \log (x \cdot y) = \log (x) + \log (y)$, then surely $\displaystyle \log (n!) = \sum\limits_{k = 1}^n {\log (k)}$.