Prove that for all positive integers n > 1,

$\displaystyle \sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=0$

My attempt:

$\displaystyle (1+x)^n=\sum_{k=0}^n\dbinom{n}{k}x^k$

Integrating both sides,

$\displaystyle \frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+C$

Substituting x = 0,

$\displaystyle C=\frac{1}{n+1}$

$\displaystyle \frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\dbinom{n}{k} \frac{x^{k+1}}{k+1}+\frac{1}{n+1}$

Substituting x = -1,

$\displaystyle 0=\sum_{k=0}^n\dbinom{n}{k} \frac{(-1)^{k+1}}{k+1}+\frac{1}{n+1}$

$\displaystyle \sum_{k=0}^n\frac{1}{k+1}\dbinom{n}{k}(-1)^{k+1}=-\frac{1}{n+1}$

What am I doing wrong?