The correct option is D √(3x^2 5x+6)+c Consider, I=∫(6x 5) dx2√(3x^2 5x+6) N ( x )=6x 5 D ( x )=3x^2 5x+6 N ( x )=D' ( x ) So, ...

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Special Integrals - 1

∫6x-5/2√3x2-5...

Question

$\int \frac{(6 x - 5)}{2 \sqrt{3 x^{2} - 5 x + 6}} d x =$

\frac{1}{\sqrt{2}} \sqrt{3 x^{2} - 5 x + 6} + c

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\sqrt{3 x^{2} - 5 x + 6} + c

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\frac{1}{2} \sqrt{3 x^{2} - 5 x + 6} + c

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\frac{1}{3} \sqrt{3 x^{2} - 5 x + 6} + c

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Solution

The correct option is D $\sqrt{3 x^{2} - 5 x + 6} + c$
Consider, $I = \int \frac{(6 x - 5) d x}{2 \sqrt{3 x^{2} - 5 x + 6}}$
$N (x) = 6 x - 5$
$D (x) = 3 x^{2} - 5 x + 6$
$N (x) = D^{'} (x)$
So, $\frac{1}{2} \int \frac{N (x)}{\sqrt{D} (x)} . d x = \frac{1}{2} \int \frac{D^{1} (x) . d x}{\sqrt{D (x)}}$
$= \frac{1}{2} \int \frac{d (D (x))}{\sqrt{D (x)}}$
$= \frac{1}{2} \times 2 \sqrt{D (x)} + c$
$= \sqrt{D (x)} + c$
$D (x) = 3 x^{2} - 5 x + 6$
So,
$I = \sqrt{3 x^{2} - 5 x + 6} + c$
Hence, $\int \frac{(6 x - 5)}{2 \sqrt{3 x^{2} - 5 x + 6}} . d x = \sqrt{3 x^{2} - 5 x + 6} + c$

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