x+3x^2+ 4x+5 gt; 1 here x^2 + 4x + 5 gt; 0 ∀ xϵ R because D lt; 0 x+3 #160; gt; x^2 + 4x + 5 x^2 + 3x + 2 lt; 0 (x ...

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Integration by Parts

Solve: x+3x...

Question

Solve: $\frac{x + 3}{x^{2} + 4 x + 5} > 1$

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Solution

$\frac{x + 3}{x^{2} + 4 x + 5} > 1$

here $x^{2} + 4 x + 5 > 0 \forall x ϵ R$
because $D < 0$
$x + 3 > x^{2} + 4 x + 5$
$x^{2} + 3 x + 2 < 0$
$(x + 2) (x + 1) < 0$

$\Rightarrow x ϵ (- 2, - 1)$ Answer

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x^{2} + 2 | x - 3 | - 10 < 0,

\frac{∣ ∣ x^{2} - 4 x ∣ ∣ + 3}{x^{2} + | x - 5 |} \geq 1

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| x^{2} - 4 x + 3 | = | x^{2} - 2 x + 1 |

\frac{x - 1}{x^{2} - 4 x + 3} < 1

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