∫x−1x+1⋅1xx2+x+1dx=

∫x−1x+11xx2+x+1dx=∫x2−1(x+1)2x⋅xx+1+1x=∫x2−1x2+2x+1xx+1+1x=∫x2−1xx+2+1xxx+1+1x=∫1−1/x2x+1x+2x+1x+1=∫2t dtt2+1t=2tan−1⁡t+c=2tan−1⁡x+1x+1+C

∫x−1x+1⋅1xx2+x+1dx=

∫x−1x+11xx2+x+1dx=∫x2−1(x+1)2x⋅xx+1+1x=∫x2−1x2+2x+1xx+1+1x=∫x2−1xx+2+1xxx+1+1x=∫1−1/x2x+1x+2x+1x+1=∫2t dtt2+1t=2tan−1⁡t+c=2tan−1⁡x+1x+1+C

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$\int \frac{x - 1}{x + 1} \cdot \frac{1}{\sqrt{x (x^{2} + x + 1)}} d x =$

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

$\sqrt{2} \tan^{- 1} \sqrt{x + \frac{1}{x} + 1} + C$

$2 \tan^{- 1} \sqrt{x + \frac{1}{x} + 1} + C$

$\frac{1}{\sqrt{2}} \tan^{- 1} \sqrt{x + \frac{1}{x} + 1} + C$

answer is C.

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Detailed Solution

$\int \frac{x - 1}{x + 1} \frac{1}{\sqrt{x (x^{2} + x + 1)}} dx$

$= \int \frac{x^{2} - 1}{(x + 1)^{2} \sqrt{x \cdot x (x + 1 + \frac{1}{x})}}$

$\begin{array}{l} = \int \frac{x^{2} - 1}{(x^{2} + 2 x + 1) x \sqrt{x + 1 + \frac{1}{x}}} \\ = \int \frac{x^{2} - 1}{x (x + 2 + \frac{1}{x}) x \sqrt{x + 1 + \frac{1}{x}}} \\ = \int \frac{1 - 1 / x^{2}}{x + \frac{1}{x} + 2) \sqrt{(x + \frac{1}{x} + 1)}} \end{array}$