1-5-4x-2x-2-dx-

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

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

Commented by abdomathmax last updated on 27/Mar/20

I = \int \frac{d x}{\sqrt{- 2 x^{2} - 4 x + 5}} w e h a v e

I = \int \frac{d x}{\sqrt{5 - 2 (x^{2} + 2 x + 1 - 1)}} = \int \frac{d x}{\sqrt{5 - 2 {(x + 1)}^{2} + 2}}

= \int \frac{d x}{\sqrt{7 - 2 {(x + 1)}^{2}}} = \int \frac{d x}{\sqrt{7} \times \sqrt{1 - \frac{2}{7} {(x + 1)}^{2}}}

v h a n g e m e n t \sqrt{\frac{2}{7}} (x + 1) = u g i v e

I = \frac{1}{\sqrt{7}} \int \frac{1}{\sqrt{1 - u^{2}}} \times \frac{\sqrt{7}}{\sqrt{2}} d u

= \frac{1}{\sqrt{2}} a r c s i n (u) + C

= \frac{1}{\sqrt{2}} a r c s i n (\sqrt{\frac{2}{7}} (x + 1)) + C

Answered by jagoll last updated on 27/Mar/20

1 + 4 - 4 x - {2 x}^{2} = 1 + 2 (2 - 2 x - x^{2})

1 + 2 (2 - (x^{2} + 2 x + 1)) =

1 + 2 (3 - {(x + 1)}^{2}) = 7 - 2 {(x + 1)}^{2}

let x + 1 = \sqrt{\frac{7}{2}} \sin t

\int \frac{\sqrt{\frac{7}{2}} \cos t dt}{\sqrt{7 - 7 \sin^{2} t}} =

\frac{\sqrt{2}}{2} \int dt = \frac{1}{2} t + c

\frac{\sqrt{2}}{2} arc \sin (\frac{\sqrt{2}}{\sqrt{7}} (x + 1)) + c

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