The value of ∫01 ∏r=1n (x+r)∑k=1n 1x+kdx is

Let ∫01 ∏r=1n (x+r)∑k=1n 1x+kdxthe given integrand is differential coefficient of ∏r=1n (x+r)⇒I=∏r=1n (x+r)01=(n+1)!−n!=n⋅n!Alternate MethodConsider  (x+1)(x+2)(x+3)⋯(x+n)=etSo that when x = 0, then t = log (n!)and when x = 1, then t = log (n + 1)!log (x+1)+ log (x +2)+ log (x+ 3)+……. + log (x + n) =t∴ 1x+1+1x+2+1x+3+⋯+1x+ndx=dt∴I=∫ln⁡n!ln⁡(n+1)! etdt=etln⁡n!ln⁡(n+1)! =eln⁡(n+1)!−eln⁡n!=(n+1)!−n!=n⋅n!