If ∫x2−x+1x2+1ecot−1 x dx=A(x)ecot−1x+c, then Ax is equal to

∫x2-x+1x2+1ecot-1xdx =∫ecot-1xx2+1x2+1-xx2+1dx =∫ecot-1x-1x2+1x+1dx =∫eg(x)g'x f(x)+f'(x)dx =eg(x)f(x)+c where g(x)=cot-1x =ecot-1x x+c f(x)=x ∴A(x)=x

If ∫x2−x+1x2+1ecot−1 x dx=A(x)ecot−1x+c, then Ax is equal to

∫x2-x+1x2+1ecot-1xdx =∫ecot-1xx2+1x2+1-xx2+1dx =∫ecot-1x-1x2+1x+1dx =∫eg(x)g'x f(x)+f'(x)dx =eg(x)f(x)+c where g(x)=cot-1x =ecot-1x x+c f(x)=x ∴A(x)=x

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If $\int \frac{x^{2} - x + 1}{x^{2} + 1} e^{\cot^{- 1}} x d x = A (x) e^{\cot^{- 1} x} + c,$ then $A (x)$ is equal to

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$- x$

$x$

$\sqrt{1 - x}$

$\sqrt{1 + x}$

answer is B.

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Detailed Solution

$\int \frac{x^{2} - x + 1}{x^{2} + 1} e^{c o t^{- 1} x} d x = \int e^{c o t^{- 1} x} [\frac{x^{2} + 1}{x^{2} + 1} - \frac{x}{x^{2} + 1}] d x = \int e^{c o t^{- 1} x} [(\frac{- 1}{x^{2} + 1}) x + 1] d x = \int e^{g (x)} [g' (x) f (x) + f' (x)] d x = e^{g (x)} f (x) + c w h e r e g (x) = c o t^{- 1} x = e^{c o t^{- 1} x} x + c f (x) = x ∴ A (x) = x$