I want to compute this thing $\sum_{k=0}^{n}{\binom{n}{k}\frac{1}{2k+1}}$ without a dropped induction.

Question

I want to compute this expression without using a ' known induction to apply':

$$\sum_{k=0}^{n}{\binom{n}{k}\frac{1}{2k+1}}$$

I saw this one :Is the numerator of $\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}$ always a power of $2$ in lowest terms?

First proof is by induction, the second one uses hypergeometric function which is a bit to non-elementary.

Thanks.

(I tried Pascal, then some identity on binomial coeff , the sum I want to calculate is from the associated integral from zero to one)