A sweet integral

First of all, we note that the inverse of $\sqrt{\log _2{(x+1)}}$ is $2^{x^2}-1$

Let $\sqrt{\log _2{(x+1)}} = f(x)$

Then we can rewrite the integral as $\int_{0}^{1} (f(x)+f^{-1}(x)+1) dx$ .

Next, note that for any invertible $f(x)$ , $\int_{a}^{b} f(x) dx+ \int_{f(a)}^{f(b)}f^{-1}(x) dx = f(b).b-f(a).a$

(To visualize this result, try plotting a graph of any invertible function and think of the integral in terms of area)

Finally, we note that $f(0)=f^{-1}(0)=0$ and $f(1)=f^{-1}(1)=1$

Plugging in the values, we get the answer 2.