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?

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. 
