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.
