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Question
\[\frac{2x - 1}{x^2 + 1}\]
Solution
\[\text{ Let u } = 2x - 1; v = x^2 + 1; \]
\[\text{ Then }, u' = 2; v' = 2x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{2x - 1}{x^2 + 1} \right) = \frac{\left( x^2 + 1 \right)2 - \left( 2x - 1 \right)2x}{( x^2 + 1 )^2}\]
\[ = \frac{2 x^2 + 2 - 4 x^2 + 2x}{( x^2 + 1 )^2}\]
\[ = \frac{- 2 x^2 + 2x + 2}{( x^2 + 1 )^2}\]
\[ = \frac{2\left( 1 + x - x^2 \right)}{( x^2 + 1 )^2}\]
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