You should solve the given improper integral such that:

`int_2^oo 1/(x*(ln x)^p)dx = lim_(n->oo) int_2^n 1/(x*(ln x)^p)dx`

You need to solve the indefinite integral `int 1/(x*(ln x)^p)dx` using the substitution `ln x = t` such that:

`ln x = t => 1/x dx = dt`

Changing the variable yields:

`int 1/(x*(ln...

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You should solve the given improper integral such that:

`int_2^oo 1/(x*(ln x)^p)dx = lim_(n->oo) int_2^n 1/(x*(ln x)^p)dx`

You need to solve the indefinite integral `int 1/(x*(ln x)^p)dx` using the substitution `ln x = t` such that:

`ln x = t => 1/x dx = dt`

Changing the variable yields:

`int 1/(x*(ln x)^p)dx = int (dt)/(t^p)`

Using the property of negative power yields:

`1/t^p = t^(-p)`

`int (dt)/(t^p) = intt^(-p) dt =(t^(-p+1))/(-p+1) + c`

Using the fundamental theorem of calculus yields:

`int_2^n 1/(x*(ln x)^p)dx = ((ln x)^(-p+1))/(-p+1)|_2^n`

`int_2^n 1/(x*(ln x)^p)dx = ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p)`

Evaluating the limit yields:

`lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) = lim_(n->oo) ((ln n)^(1-p))/(1-p) - lim_(n->oo)((ln 2)^(1-p))/(1-p) `

If `p>1 => lim_(n->oo)((ln n)^(1-p))/(1-p) = lim_(n->oo) 1/(((ln n)^(p-1))(1-p)) = 1/oo = 0`

`lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) =0 - ((ln 2)^(1-p))/(1-p) = - ((ln 2)^(1-p))/(1-p)`

If `p<1 => lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) = oo`

**Hence, evaluating the given improper integral yields that integral converges if p>1 and integral diverges that p<1.**