Improper integrals with bounds 0 to infinity $$\int_{0}^{\infty }\frac{1}{x^{1/2}+x^{3/2}}dx$$
I managed to integrate correctly and get $2\arctan(\sqrt{x})+C$, but I was wondering why my teacher split it up into two pieces, one from $0$ to $1$ with a limit at $0$ evaluating $2\arctan(\sqrt{x})+C$ and the other $1$ to $\infty$ with a limit at infinity evaluating $2\arctan(\sqrt{x})+C$. Can I just take $2\arctan(\sqrt{x})$ and evaluate from $0$ to $\infty$ with a limit at infinity?

If not wrong: You could solve the problem but you want to know why did your teacher split the above integral as
$$\int_0^\infty = \int_0^1 + \int_1^\infty$$
or as
$$\arctan(x)|_0^\infty = \arctan(x)|_0^1 + \arctan(x)|_1^\infty$$
then Comment by @Accelerator
however, for this particular case
$$\begin{align*}\int_0^\infty \frac {dx}{x^{\frac 12} + x^{\frac 32}}
&=\left(\int_0^1 + \int_1^\infty\right)\frac {1}{x^{\frac 12} + x^{\frac 32}}dx
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_1^\infty\frac {dx}{x^{\frac 12} + x^{\frac 32}}}
; \text{ Let } \color{green}{x = \frac 1y }
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}}  + \int_1^0\frac {\frac{-dy}{y^2}}{y^{\frac {-1}2} + y^{\frac {-3}2}}
\\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}}  + \color{blue}{\int_0^1 \frac {dy}{y^{\frac 12} + y^{\frac {3}2}}}
\\& = 2\times\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}}
\\& = 2\times\int_0^1 \frac {x^{-\frac 12}dx}{1 + x}
\\& = 2\frac {\beta\left(\frac 12, \frac 12\right)}{2} = \pi
\end{align*}$$
Why did I use $\beta$ function?: Master Theorem
