Challenging convergent improper

Question

I am interested in finding an answer to $ \int_{0}^{1} \frac{2^x-1}{2x\sqrt{x}} dx $. I am not sure if there is a closed form and if it can be done in terms of elementary functions. I tried Integration by Parts but that leads me to nowhere. Putting it in Desmos reveals $0.785861$ which is incredibly close to $\pi/4$. Any insight of the math community would be appreciated. If it can't be done, I accept that fate

Answer 1

$\int\limits_0^1 \frac{2^x-1}{2x \sqrt{x}} dx= -1+\sqrt{\pi \ln 2} \;{\rm erfi}(\sqrt{\ln 2})$, where ${\rm erfi}(z)=-i \, {\rm erf}(i z)$.

Answer 2

To get rid of the fractional power, let $x=t^2$

$$I=\int \frac{2^x-1}{2x\sqrt{x}}\,dx=\int \frac{2^{t^2}-1}{t^2}\,dt=\int \frac {e^{t^2 \log(2)} }{t^2}\,dt$$ $$t \sqrt{\log(2)}=u\quad \implies \quad I=\sqrt{\log(2)}\int \frac{e^{u^2}-1}{u^2}\,du$$ which starts to look familiar.

Use a couple of integrations by parts to obtain $I$; go back to $x$ and apply the bounds.