∫2x2−1+x2x2+4x2(x2+4) dx=1

∫2x2−1+x2x2+4x2(x2+4)dx =∫2x2-1x2(x2+4)dx+∫x2x2+4x2(x2+4)dx =∫x2+(x2+4)-5x2(x2+4)+∫1x2+4dx =∫1x2+4+1x2-5x2(x2+4)+sin h-1 x2+c =12 tan-1 x2+-1x-54∫(x2+4)-x2x2(x2+4)dx+sin h-1x2+c=12tan-1x2-1x-54∫1x2-1x2+4dx+sin h-1x2+c =12tan-1x2-1x-54-1x+58tan-1x2+sin h-1x2 =98tan-1x2+14x+sin h-1x2+c

∫2x2−1+x2x2+4x2(x2+4) dx=1

∫2x2−1+x2x2+4x2(x2+4)dx =∫2x2-1x2(x2+4)dx+∫x2x2+4x2(x2+4)dx =∫x2+(x2+4)-5x2(x2+4)+∫1x2+4dx =∫1x2+4+1x2-5x2(x2+4)+sin h-1 x2+c =12 tan-1 x2+-1x-54∫(x2+4)-x2x2(x2+4)dx+sin h-1x2+c=12tan-1x2-1x-54∫1x2-1x2+4dx+sin h-1x2+c =12tan-1x2-1x-54-1x+58tan-1x2+sin h-1x2 =98tan-1x2+14x+sin h-1x2+c

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$\int \frac{2 x^{2} - 1 + x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} d x = 1$

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$\frac{9}{8} T a n^{- 1} \frac{x}{2} + \frac{1}{4 x} + \cos h^{- 1} \frac{x}{2} + c$

$\frac{9}{8} T a n^{- 1} \frac{x}{2} + \frac{1}{4 x} + \sin h^{- 1} \frac{x}{2} + c$

$\frac{9}{16} \log |\frac{x + 2}{x - 2}| + \frac{1}{4 x} + \log |\frac{x + \sqrt{x^{2} + 4}}{2}| + c$

$\frac{9}{16} \log |\frac{2 - x}{2 - x}| + \frac{1}{4 x} + \cos h^{- 1} \frac{x}{2} + c$

answer is B.

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Detailed Solution

$\int \frac{2 x^{2} - 1 + x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} dx = \int \frac{2 x^{2} - 1}{x^{2} (x^{2} + 4)} dx + \int \frac{x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} dx = \int \frac{x^{2} + (x^{2} + 4) - 5}{x^{2} (x^{2} + 4)} + \int \frac{1}{\sqrt{x^{2} + 4}} dx = \int (\frac{1}{x^{2} + 4} + \frac{1}{x^{2}} - \frac{5}{x^{2} (x^{2} + 4)}) + \sin h^{- 1} \frac{x}{2} + c = \frac{1}{2} \tan^{- 1} \frac{x}{2} + (\frac{- 1}{x}) \frac{- 5}{4} \int \frac{(x^{2} + 4) - x^{2}}{x^{2} (x^{2} + 4)} d x + \sin h^{- 1} (\frac{x}{2}) + c$

$= \frac{1}{2} \tan^{- 1} \frac{x}{2} \frac{- 1}{x} \frac{- 5}{4} \int (\frac{1}{x^{2}} - \frac{1}{x^{2} + 4}) dx + \sin h^{- 1} (\frac{x}{2}) + c = \frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{x} - \frac{5}{4} (\frac{- 1}{x}) + \frac{5}{8} \tan^{- 1} (\frac{x}{2}) + \sin h^{- 1} (\frac{x}{2}) = \frac{9}{8} \tan^{- 1} \frac{x}{2} + \frac{1}{4 x} + \sin h^{- 1} (\frac{x}{2}) + c$