$\frac{d}{dx}\left(\frac{-x^2+2x+1}{4x-3}\right)=\frac{\left(-2x+2\right)\left(4x-3\right)+4\left(x^2-2x-1\right)}{\left(4x-3\right)^2}$
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Integral
$\int\frac{-x^2+2x+1}{4x-3}dx=-\frac{1}{8}x^2+\frac{5}{16}x+\frac{31}{64}\ln\left(4x-3\right)+C_0$
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