Question

The value of ${\int }_0^1\left(\prod _{r-1}^n(x+r)\right)\left(\sum _{k=1}^n\frac{1}{x+k}\right)dx$ equals

• A. $n$
• B. $n!$
• C. $(n+1)!$
• D. $n.n!$

Answer 1

Answer: (D)

$n.n!$

$l={\int }_0^1\prod _{r=1}^n(x+r)\sum _{k=1}^n\frac{1}{x+k}dx$

Let $\ln \left(\prod _{r=1}^n(x+r)\right)=t$

$\sum _{k=1}^n\ln (x+k)=t$

$\Rightarrow {\sum }_{k=1}^n\frac{1}{x+k}dx=dt$
$l={\int }_{ln(n!)}^{ln(n+1)!}{e}^{t}dt={\left[e^t\right]}_{ln\left(n\right)!}^{ln(n+1)!}$
$(n+1)!-n!=n\cdot n!$