ddxa tan-1x+b logx-1x+1=1x4-1⇒a-2b=

y=a tan-1x+b log x-1x+1 =a tan-1x+b(log(x-1)-log(x+1)) dydx=a1+x2+b1x-1-1x+1 =a1+x2+b2x2-1 =a(x2-1)+2b(x2+1)x4-1 =x2(a+2b)+(2b-a)x4-1=1x4-1 a+2b=0, 2b-a=1 a-2b=-1

ddxa tan-1x+b logx-1x+1=1x4-1⇒a-2b=

y=a tan-1x+b log x-1x+1 =a tan-1x+b(log(x-1)-log(x+1)) dydx=a1+x2+b1x-1-1x+1 =a1+x2+b2x2-1 =a(x2-1)+2b(x2+1)x4-1 =x2(a+2b)+(2b-a)x4-1=1x4-1 a+2b=0, 2b-a=1 a-2b=-1

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$\frac{d}{dx} [a \tan^{- 1} x + b \log (\frac{x - 1}{x + 1})] = \frac{1}{x^{4}} - 1 \Rightarrow a - 2 b =$

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Detailed Solution

$y = a \tan^{- 1} x + b \log (\frac{x - 1}{x + 1}) = a \tan^{- 1} x + b (\log (x - 1) - \log (x + 1)) \frac{dy}{dx} = \frac{a}{1 + x^{2}} + b [\frac{1}{x - 1} - \frac{1}{x + 1}] = \frac{a}{1 + x^{2}} + b [\frac{2}{x^{2} - 1}] = \frac{a (x^{2} - 1) + 2 b (x^{2} + 1)}{x^{4} - 1} = \frac{x^{2} (a + 2 b) + (2 b - a)}{x^{4} - 1} = \frac{1}{x^{4} - 1} a + 2 b = 0, 2 b - a = 1 a - 2 b = - 1$