<s>[INST] integrate $\int \frac{x-3}{\sqrt{1-x^2}}$ 
$$\int \frac{x-3}{\sqrt{1-x^2}} \mathrm dx$$

I know that $\int \frac{1}{1-x^2}\mathrm dx=\arcsin(\frac{x}{1})$ but how can I continue from here? 
 [/INST] Notice, $$\int \frac{x-3}{\sqrt {1-x^2}}\ dx$$$$=\int \frac{x}{\sqrt {1-x^2}}\ dx-\int \frac{3}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\int \frac{(-2x)}{\sqrt {1-x^2}}\ dx-3\int \frac{1}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\int (1-x^2)^{-1/2}\ d(1-x^2)-3\int \frac{1}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\frac{(1-x^2)^{1/2}}{1/2}-3\sin^{-1}(x)+C$$
$$=-\sqrt{1-x^2}-3\sin^{-1}(x)+C$$
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