Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form

I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.

So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb + \binom{n}{n}x^n$$ Now on integrating both sides and putting $x=1$.

I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is $$\frac{2^{n+1}-1}{n+1}.$$ Where does that -1 term in the numerator come from?

$$\int^{1}_{0} (1+x)^n dx=\frac{(1+x)^{n+1}}{n+1}\bigg|^{1}_{0}=\frac{(1+1)^{n+1}-(1+0)^{n+1}}{n+1}=\frac{2^{n+1}-1}{n+1}$$

With indefinite integrals, there is always a constant of integration walking around. Hence, we use definite integrals so that the equality is kept.

Note that $$\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$$

Therefore,

\begin{align} \sum_{k=0}^{n}\frac{1}{k+1}{n\choose k}&=\frac{1}{n+1}\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k} \\&=\frac{1}{n+1}\sum_{k=0}^{n}{{n+1}\choose {k+1}} \\&=\frac{1}{n+1}\left [\sum_{k=0}^{n+1}{{n+1}\choose k}-1\right ] \\&=\frac{1}{n+1}\left (2^{n+1}-1\right ) \end{align}