- #1

Dell

- 590

- 0

[tex]\int[/tex]dx/ln(x) (from 0-1)

what i did was look for a similar function

i took 1/x which is bigger than 1/lnx,

but 1/x diverges from 0-1

can i split my function up and compare

[tex]\int[/tex]dx/ln(x) (from 0-0.5) + [tex]\int[/tex]dx/ln(x) (from 0.5-1)

now i know that 1/x converges from 0.5-1 so 1/lnx does as well

now i need to find a fuction to compare 1/lnx with from 0-0.5

1/lnx < 1/(x+1)

1/(x+1) converges

therefore 1/lnx does as well

therefore 1/lnx is made up of two converging parts from 0-1 so the whole also converges

is this correct?