How you calculate this? $int_1^ndx/(x+|x|)$

Question

854 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T15:59:17

$int_1^ndx/(x+floor(x)) =log_e(prod_(k=1)^(n-1)(2k+1)/(2k))$

$int_1^ndx/(x+floor(x)) = sum_(k=1)^(n-1)int_k^(k+1)(dx)/(x+k) = sum_(k=1)^(n-1)log_e((2k+1)/(2k))$

or

$int_1^ndx/(x+floor(x)) =log_e(prod_(k=1)^(n-1)(2k+1)/(2k))$

How you calculate this? int_1^ndx/(x+|__x__|)int_1^ndx/(x+|__x__|)