Question

How do you use the integral test to determine whether `int dx/(x+lnx)` converges or diverges from `[2,oo)`?

Answer 1

The integral:

`int_2^oo dx/(x+lnx)`

diverges.

Explanation:

As in the interval `x in [2,+oo)` the function:

`f(x) = 1/(x+lnx)`

is positive and decreasing, and:

`lim_(x->oo) f(x) = 0`

`f(n) = 1/(n+lnn)`

based on the integral test the convergence of the improper integral:

`int_2^oo dx/(x+lnx)`

is equivalent to the convergence of the series:

`sum_(n=2)^oo 1/(n+lnn)`

Consider now the harmonic series:

`sum_(n=0)^oo 1/n = oo`

that we know to be divergent and apply the limit comparison test:

`lim_(n->oo) (1/(n+lnn))/(1/n) = lim_(n->oo) (1/(1+lnn/n)) = 1`

as the limit is finite the two series have the same character, then also the series

`sum_(n=2)^oo 1/(n+lnn)`

is divergent.