Let $y' (x) . g (x) - y (x) . g' (x) + y^{2} (x) = 0, y (- 1) = 1, x \in R$ where $f' (x)$ denotes $\frac{d f (x)}{d x}$ and $g (x)$ is a given non-constant differentiable function on R with $g (- 1) = 0$ then value of $\int_{1}^{2} (\frac{g (x)}{y (x)} . \frac{1}{x^{2} \sqrt{x^{2} + {(\frac{g (x)}{y (x)})}^{2}}}) d x =$

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$2 \sqrt{5} - \sqrt{13}$

$\frac{7}{2 (2 \sqrt{5} + \sqrt{13})}$

$\sqrt{13} - \sqrt{5}$

$\frac{7}{2 (2 \sqrt{5} - \sqrt{13})}$

answer is B.

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Detailed Solution

$y' (x) . g (x) - y (x) . g' (x) + y^{2} (x) = 0 \Rightarrow \frac{g (x)}{y (x)} = x + c = - d (\frac{g (x)}{y (x)}) + 1 = 0 y (- 1) = 1 a n d g (- 1) = 0 = \int_{1}^{2} \frac{(1 + x) d x}{x^{2} \sqrt{x^{2} {(1 + x)}^{2}}} (- \frac{2}{x^{2}} - \frac{2}{x^{3}}) d x = 2 t d t = \int_{1}^{2} \frac{(1 + x) d x}{x^{2} \sqrt{2 + \frac{2}{x} + \frac{1}{x^{2}}}} p u t 2 + \frac{2}{x} + \frac{1}{x^{2}} = t^{2} = \sqrt{5} - \frac{\sqrt{13}}{2} = \frac{2 \sqrt{5} - \sqrt{13}}{2} = \frac{7}{2 (2 \sqrt{5} + \sqrt{13})}$