Improper integrals with bounds 0 to infinity

Question

$$\int_{0}^{\infty }\frac{1}{x^{1/2}+x^{3/2}}dx$$ 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?

Answer

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 $$\arctan(x)|_0^\infty = \arctan(x)|_0^1 + \arctan(x)|_1^\infty$$ however, for this particular case $$\begin{align*}\int_0^\infty \frac {dx}{x^{\frac 12} + x^{\frac 32}} &=\left(\int_0^1 + \int_1^\infty\right)\frac {1}{x^{\frac 12} + x^{\frac 32}}dx \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_1^\infty\frac {dx}{x^{\frac 12} + x^{\frac 32}}} ; \text{ Let } \color{green}{x = \frac 1y } \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \int_1^0\frac {\frac{-dy}{y^2}}{y^{\frac {-1}2} + y^{\frac {-3}2}} \\&=\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} + \color{blue}{\int_0^1 \frac {dy}{y^{\frac 12} + y^{\frac {3}2}}} \\& = 2\times\int_0^1 \frac {dx}{x^{\frac 12} + x^{\frac 32}} \\& = 2\times\int_0^1 \frac {x^{-\frac 12}dx}{1 + x} \\& = \boxed{2\frac {\beta\left(\frac 12, \frac 12\right)}{2} = \pi} \end{align*}$$ Why did I use 𝛽 function?: Master Theorem