Question

How to calculate this? `int_(1/(n+2))^(1/n)[1/x]dx`.

Answer 1

See the answer below:

Answer 2

For `I_n = int_(1/(n+2))^(1/n) floor(1/x) dx`

`I_n= 1/(n+1)+1/(n+2)`

Explanation:

Changing variables

`y = 1/x` we have

`int_(1/(n+2))^(1/n)floor (1/x)dx equiv int_n^(n+2) floor(y) /y^2 dy` or

`int_n^(n+1)n/y^2 dy + int_(n+1)^(n+2)(n+1)/y^2 dy= 1/(n+1)+1/(n+2)`

Answer 3

Now for `I_n=int_(1/(n+2))^(1/n) [1/x] dx` is

`I_n= (8 (n+1))/(4 n (n+2)+3)`

Explanation:

Changing variables

`y = 1/x` we have

`int_(1/(n+2))^(1/n) [1/x] dx equiv int_n^(n+2) [[y]] /y^2 dy` or

`int_n^(n+1/2)n/y^2 dy + int_(n+1/2)^(n+3/2)(n+1)/y^2 dy+int_(n+3/2)^(n+2)(n+2)/y^2 dy = 1/(3 + 2 n) + (4 (1 + n))/((1 + 2 n) (3 + 2 n)) + n/(n + 2 n^2) = (8 (1 + n))/(3 + 4 n (2 + n))`