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

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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__|)#