X2-x-1-x3-8-A-x-2-Bx-C-x2-2x-4- A-B-

The correct option is B 1 LHS=x^2 x 1x^3 8; #160; #10;RHS=Ax 2+Bx+Cx^2+2x+4 #10; =A(x^2+2x+4)x^3 8+(Bx+C)(x 2) #10; x^2 x 1=A(x^2+2 ...

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Differentiation of Inverse Trigonometric Functions

x2-x-1/x3-8=A...

Question

$\frac{x^{2} - x - 1}{x^{3} - 8} = \frac{A}{x - 2} + \frac{B x + C}{x^{2} + 2 x + 4} \Rightarrow A + B =$

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\frac{x^{2} - x - 3}{x^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1} \Rightarrow 2 A + B + C =

\frac{2 x}{x^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1}

\frac{2 b^{2} + x^{2}}{b^{3} - x^{3}} - \frac{2 x}{b x + b^{2} + x^{2}} + \frac{1}{x - b} = 0

\frac{x^{2} - x + 3}{x^{3} - 1} = \frac{A}{(x - 1)} + \frac{B x + C}{(x^{2} + x + 1)}

A =

B =

C =

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

0 < | x | < \sqrt{2}

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