The value of f∫1+xxdx, is

Let I=∫1+xxdx⇒ I=∫t2t2−1−2tdt, where 1+x=t2⇒ t=2∫t2−1+1t2−1dt=2∫1+1t2−1dt⇒ I=2t+log⁡t−1t+1+C=21+x+log⁡1+x−11+x+1+C

The value of f∫1+xxdx, is

Let I=∫1+xxdx⇒ I=∫t2t2−1−2tdt, where 1+x=t2⇒ t=2∫t2−1+1t2−1dt=2∫1+1t2−1dt⇒ I=2t+log⁡t−1t+1+C=21+x+log⁡1+x−11+x+1+C

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The value of $f \int \frac{\sqrt{1 + x}}{x} d x$ , is

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$2 \sqrt{1 + x} + \log |\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}| + C$

$\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1} + C$

$2 \sqrt{1 + x} + C$

$\log_{e} |\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}| + C$

answer is A.

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Detailed Solution

Let

$I = \int \frac{\sqrt{1 + x}}{x} d x$

$\Rightarrow I = \int \frac{\sqrt{t^{2}}}{t^{2} - 1} - 2 t d t,$ where $1 + x = t^{2}$

$\begin{array}{l} \Rightarrow t = 2 \int \frac{t^{2} - 1 + 1}{t^{2} - 1} d t = 2 \int (1 + \frac{1}{t^{2} - 1}) d t \\ \Rightarrow I = 2 t + \log |\frac{t - 1}{t + 1}| + C = 2 \sqrt{1 + x} + \log |\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}| + C \end{array}$