Dividing numerator and denominator of integrand in definite integral I'm trying to solve the following definite integral $$\int_0^{\frac{\pi}{4}} \sqrt{\tan  x}\,dx$$
First, I made the substitution $u = \sqrt{\tan x}$
And arrived with: $$\int_0^1 \frac{2u^2}{u^4+1}\,du$$
When divided both numerator and denominator of the integrand by $u^2$ and with some manipulation I got
$$\int_0^1 \frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2+2}+\frac{1-\frac{1}{u^2}}{(u+\frac{1}{u})^2-2}\, du$$
This integral could be solved easily with u sub, giving the result:
$$\left.\frac{1}{2\sqrt{2}} \ln\left|\frac{u^2-\sqrt{2}u+1}{u^2+\sqrt{2}u+1}\right|+\frac{1}{\sqrt{2}}\arctan\left(\frac{1}{\sqrt{2}}\left(u-\frac{1}{u}\right)\right)\ \right|_0^1$$
The integral however is not defined at $u=0$, my question is: Are we allowed to do this kind of manipulation to definite integrals with rational function, and will proceeding with improper integral yield the correct result?

Previous answers did not address your question of whether your method is invalid due to division by zero. That is what I will address here.
There are two ways to evade the problem. The cheat way is to observe that meromorphic functions are closed under division except dividing by the everywhere-zero function, and we eliminate removable singularities. In this setting, multiplication by $u/u$ in your integral computation would not be a problem, because it is equivalent to multiplying by the constant-one function. You would also have to use the fact that if a meromorphic function $f$ on a domain $D ⊆ ℂ$ has an anti-derivative $g$, then $g{↾}(D{∖}S)$ is the unique anti-derivative of $f{↾}(D{∖}S)$ for any set $S$ of isolated points. So it does not matter if you cut off some isolated points and find the anti-derivative, because you can then extend it back (if the original function had an anti-derivative). This general fact is why you will never really encounter a problem with division by zero for sufficiently nice functions, even if the method as stated is technically wrong.
A more concrete way is to observe that $\int_0^1 f(u)\ du = \lim_{t→0^+} \int_t^1 f(u)\ du$ if the former exists, and so once you find $\int_t^1 f(u)\ du = g(t)$ for every $t∈ℝ^+$, you can then take limits, no problem.
