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I have simple integral that must be solved by substitution. $$\int \frac{1}{\sqrt{1-2x-x^2}} \, dx = \int \frac{1}{\sqrt{2-(x+1)^2}} \, dx $$ After substitution $$u = x+1 $$ we get $$\int \frac{1}{\sqrt{-u^2+2}} \, du $$ But i dont know how to continue. Thanks for any help.

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Recall that

$$(\arcsin x)'=\frac1{\sqrt{1-x^2}}$$ and adapt to the given integral.

Use the change of variable $u=\sqrt2x$.

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Substitute $$\sqrt{1-2x-x^2}=tx+1$$ it is the so-called Euler substitution.

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  • $\begingroup$ Thanks going to learn about that euler subs. $\endgroup$ – Juraj Jakubov Mar 9 at 17:28
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Try $\frac{u}{\sqrt{2}}=\cos({t})$ and use $\cos^2(t)+\sin^2(t)=1$.

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Now put $u=\sqrt{2}\sin\theta$

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