$\frac{2 x}{{2 x}^{2} + 5 x + 2} > \frac{1}{(x + 1)}$ . Find the number of integral solutions for the given inequality.

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Solution

$\frac{2 x}{{2 x}^{2} + 5 x + 2} > \frac{1}{(x + 1)}$
$\Rightarrow \frac{2 x}{{2 x}^{2} + 5 x + 2} - \frac{1}{(x + 1)} > 0$
$\Rightarrow \frac{2 x (x + 1) - {2 x}^{2} - 5 x - 2}{{(2 x}^{2} + 5 x + 2) (x + 1)} > 0$
$\Rightarrow \frac{- 3 x - 2}{{(2 x}^{2} + 5 x + 2) (x + 1)} > 0$
$\Rightarrow \frac{- (3 x + 2)}{{(2 x}^{2} + 4 x + x + 2) (x + 1)} > 0$
$\Rightarrow \frac{- (3 x + 2)}{(2 x + 1) (x + 2) (x + 1)} > 0$
$\Rightarrow x \in (- 2, - 1) \cup (- \frac{2}{3}, - \frac{1}{2})$
Therefore, number of integral solutions $= 0$ .

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