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?
On
Another solution:
Use a trigonometric substitution $x=\tan^2 u$ and $dx=2\tan u \sec^2 u dx$
Thus:
$$\displaystyle\int \frac{dx}{\sqrt{x+1}}= \int \frac{2\tan u \sec^2 u \,du}{\sqrt{\tan^2 u+1}}=2\int \tan u \sec u \,du=2 \sec u +C=2 \sqrt{x+1}+C, $$
as expected.
Please note that I know this does not answer the OP's question. I merely wanted to present another method of solution.
Yes it is correct
$$\require{cancel}\int\frac{2\cancel{u}}{\cancel u}du=\int 2 du=2u+\mathcal C\overset{u=\sqrt{x+1}}{=}2\sqrt{x+1}+\mathcal C$$