Substitution Issue In the integral $\int \frac{dx}{\sqrt{x+1}}$, substituting $u = \sqrt{x+1}$ and $du = \frac{1}{2\sqrt{x+1}}dx$ yields $\int \frac{2\sqrt{x+1}}{u}du$, but since $u = \sqrt{x+1}$ then the integral is just $\int\frac{2u}{u}du$ which is incorrect. 
How is it incorrect?

Yes it is correct
$$\require{cancel}\int\frac{2\cancel{u}}{\cancel u}du=\int 2 du=2u+\mathcal C\overset{u=\sqrt{x+1}}{=}2\sqrt{x+1}+\mathcal C$$
