Is the summation $\sum_{i=1}^{n}\frac1{i} \binom{n}{i}$ possible?

To expand a little on @Sangchul Lee's comment - after the substitution of ${y=1+x}$ the integral becomes

$${\Rightarrow \int_{1}^{2}\frac{y^n-1}{y-1}dy}$$

(This substitution isn't "necessary" for the next part, however it makes it a bit clearer). Now we can use a special factoring formula:

$${a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})}$$

To get

$${\int_{1}^{2}\frac{(y-1)(y^{n-1} + y^{n-2} + ... + y + 1)}{(y-1)}dy=\int_{1}^{2}y^{n-1} + y^{n-2} + ... + y + 1dy}$$

Evaluating that last integral gives us the sum

$${\Rightarrow \sum_{k=1}^{n}\frac{2^{k}-1}{k}}$$

So overall you have that

$${\sum_{k=1}^{n}\frac{{n\choose k}}{k}=\sum_{k=1}^{n}\frac{2^k-1}{k}}$$

Other than that though, I'm not sure if there's any more useful form. WolframAlpha gives some very nasty looking closed forms involving special functions.