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Question Number 168138 by Shemson_buo last updated on 04/Apr/22

Answered by MJS_new last updated on 04/Apr/22

$$\int \frac{dx}{\sqrt{1-2x-x^2}}=$$$$[t=\arcsin \frac{x+1}{\sqrt{2}}\to dx=\sqrt{1-2x-x^2}dt]$$$$=\int dt=t=\arcsin \frac{x+1}{\sqrt{2}}+C$$

Commented by MJS_new last updated on 04/Apr/22

$$\mathrm{in}3\mathrm{steps}$$$$\int \frac{dx}{\sqrt{1-2x-x^2}}=$$$$[r=x+1\to dx=dr]$$$$=\int \frac{dr}{\sqrt{2-r^2}}=$$$$[s=\frac{r}{\sqrt{2}}\to dr=\sqrt{2}ds]$$$$=\int \frac{ds}{\sqrt{1-s^2}}=$$$$[t=\arcsin s\to ds=\sqrt{1-s^2}dt]$$$$=\int dt=t=\arcsin s=\arcsin \frac{r}{\sqrt{2}}=$$$$=\arcsin \frac{x+1}{\sqrt{2}}+C$$

Answered by Mathspace last updated on 04/Apr/22

$$I=\int \frac{dx}{\sqrt{1-2x-x^2}}$$$$wehave1-2x-x^2=-(x^2+2x-1)$$$$=-(x^2+2x+1-2)$$$$=2-{(x+1)}^{2}andchangement$$$$x+1=\sqrt{2}sintgive$$$$I=\int \frac{\sqrt{2}costdt}{\sqrt{2}cost}=\int dt+C$$$$=t+C=arcsin\left(\frac{x+1}{\sqrt{2}}\right)+C$$

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