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Question Number 86092 by ar247 last updated on 27/Mar/20

$$\int \frac{1}{\sqrt{5-4x-2x^2}}dx$$

Commented by abdomathmax last updated on 27/Mar/20

$$I=\int \frac{dx}{\sqrt{-2x^2-4x+5}}wehave$$$$I=\int \frac{dx}{\sqrt{5-2(x^2+2x+1-1)}}=\int \frac{dx}{\sqrt{5-2{(x+1)}^2+2}}$$$$=\int \frac{dx}{\sqrt{7-2{(x+1)}^2}}=\int \frac{dx}{\sqrt{7}\times \sqrt{1-\frac{2}{7}{(x+1)}^2}}$$$$vhangement\sqrt{\frac{2}{7}}(x+1)=ugive$$$$I=\frac{1}{\sqrt{7}}\int \frac{1}{\sqrt{1-u^2}}\times \frac{\sqrt{7}}{\sqrt{2}}du$$$$=\frac{1}{\sqrt{2}}arcsin\left(u\right)+C$$$$=\frac{1}{\sqrt{2}}arcsin\left(\sqrt{\frac{2}{7}}(x+1)\right)+C$$

Answered by jagoll last updated on 27/Mar/20

$$1+4-4\mathrm{x}-{2\mathrm{x}}^2=1+2(2-2\mathrm{x}-{\mathrm{x}}^2)$$$$1+2(2-({\mathrm{x}}^2+2\mathrm{x}+1))=$$$$1+2(3-{(\mathrm{x}+1)}^2)=7-2{(\mathrm{x}+1)}^2$$$$\mathrm{let}\mathrm{x}+1=\sqrt{\frac{7}{2}}\sin \mathrm{t}$$$$\int \frac{\sqrt{\frac{7}{2}}\cos \mathrm{t}\mathrm{dt}}{\sqrt{7-7\sin {}^2\mathrm{t}}}=$$$$\frac{\sqrt{2}}{2}\int \mathrm{dt}=\frac{1}{2}\mathrm{t}+\mathrm{c}$$$$\frac{\sqrt{2}}{2}\mathrm{arc}\sin \left(\frac{\sqrt{2}}{\sqrt{7}}(\mathrm{x}+1)\right)+\mathrm{c}$$$$$$

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