Solution of dy-dx-y2-y-1-x2-x-1-0 is-

The correct option is A tan ^ 1 ( 2x+1 /√( 3 )) +tan ^ 1 ( 2y+1 /√( 3 ) )=c ⇒ #160; dy(y+12)^2+(√(3)2)^2 = dx(x+12)^2+ (√(3)2)^2 ⇒ #16 ...

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Solution of $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ is:

{tan}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + {tan}^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = c

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{sin}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + {sin}^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = c

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{sinh}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + {sinh}^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = c

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{sin}^{- 1} (\frac{2 x + 1}{\sqrt{3}}) = c

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{cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + {sin}^{- 1} (\frac{2 x}{x^{2} + 1}) + {tan}^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

y = {sin}^{- 1} \frac{2 x}{1 + x^{2}} + {tan}^{- 1} x

\frac{d y}{d x}

{cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) + {tan}^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3},

s i n^{- 1} (\frac{2 x}{1 + x^{2}})

t a n^{- 1} (\frac{2 x}{1 - x^{2}})

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