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$$\int \frac{x-3}{\sqrt{1-x^2}} \mathrm dx$$

I know that $\int \frac{1}{1-x^2}\mathrm dx=\arcsin(\frac{x}{1})$ but how can I continue from here?

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    $\begingroup$ Split the integral into two parts. The first part can be done by substitution. $\endgroup$ – Anurag A Jan 13 '16 at 13:20
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Write $\int \frac{x-3}{\sqrt{1-x^2}}\mathrm dx=\int \frac{x}{\sqrt{1-x^2}}\mathrm dx-\int \frac{3}{\sqrt{1-x^2}} \mathrm dx$.

For the first term, let $u=1-x^2$, leading to $\mathrm du=-2x \mathrm dx$. If you're still having trouble, write $\frac{-du}{2}=x \mathrm dx$, which appears in the numerator of your first term.

For the second term, you're on the correct path, but it should be $\int \frac{1}{\sqrt{1-x^2}} \mathrm dx=\arcsin x+C$

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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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Two hints:

  • $$\frac{x-3}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}} - \frac{3}{\sqrt{1-x^2}}$$
  • You had it a bit wrong: $$\int\frac{1}{\sqrt{1-x^2}} = \arcsin(x)$$
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Write $\int \frac {x-3}{\sqrt {1-x^2}} = \int \frac {x}{\sqrt {1-x^2}}-3\int \frac 1{\sqrt {1-x^2}}$.

Use the substitution $1-x^2=u$ in the first integral of the RHS. Also you know the second integral of the RHS as per your post.

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