If $\int \frac{d x}{{(x + \sqrt{x (1 + x)})}^{2}} = 2 (f (x) + g (x)) + C$ where C is constant of integration. Given $f (- 1) = 0, g (- 1) = - 1$ then which of the following is/are TRUE?

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$\lim_{x \to \infty} f (x) = \ln (\frac{1}{2})$

$\lim_{x \to \infty} g (x) = 0$

$\lim_{x \to \infty} f (x) = \ln 2$

$\lim_{x \to \infty} g (x) = 1$

answer is A.

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Detailed Solution

$\int \frac{d x}{{(x + \sqrt{x (1 + x)})}^{2}} = \int \frac{d x}{{(\sqrt{x^{2} + x} + x)}^{2}} c o n s i d e r x + \sqrt{x^{2} + x} = t \sqrt{x^{2} + x} = t - x x^{2} + x = t^{2} + x^{2} - 2 t x \Rightarrow (2 t + 1) x = t^{2} \Rightarrow x = \frac{t^{2}}{2 t + 1} d x = (\frac{(2 t + 1) (2 t) - t^{2} (2)}{{(2 t + 1)}^{2}}) d t = \frac{2 t^{2} + 2 t}{{(2 t + 1)}^{2}} d t \int \frac{(\frac{2 t^{2} + 2 t}{{(2 t + 1)}^{2}}) d t}{t^{2}} = \int \frac{2 d t}{{(2 t + 1)}^{2}} + \int \frac{2}{t {(2 t + 1)}^{2}} d t = 2 \ln (\frac{t}{2 t + 1}) + \frac{1}{2 t + 1} + C, t = x + \sqrt{x^{2} + x}$