Question

1. $\frac{dy}{dx}=\frac{1}{\sqrt{1 - x^{2}}}-\frac{1}{\sqrt{x}}$

Explanation:

Step1: Integrate each term separately

$$y=\int\frac{1}{\sqrt{1 - x^{2}}}dx-\int\frac{1}{\sqrt{x}}dx$$

Step2: Apply integration formulas

For $\int\frac{1}{\sqrt{1 - x^{2}}}dx$, by the formula $\int\frac{1}{\sqrt{1 - t^{2}}}dt=\arcsin(t)+C_1$, here $t = x$, so $\int\frac{1}{\sqrt{1 - x^{2}}}dx=\arcsin(x)+C_1$.

For $\int\frac{1}{\sqrt{x}}dx=\int x^{-\frac{1}{2}}dx$. Using the power - rule $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C_2(n
eq - 1)$, with $n=-\frac{1}{2}$, we have $\int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_2=\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C_2 = 2\sqrt{x}+C_2$.

Step3: Combine the results

$$y=\arcsin(x)-2\sqrt{x}+C$$ (where $C = C_1 - C_2$)

Answer:

$y=\arcsin(x)-2\sqrt{x}+C$