∫2x2−1+x2x2+4x2(x2+4) dx=

∫2x2−1+x2x2+4x2(x2+4) dx 2x2−1+x2x2+4x2(x2+4)=2x2−1x2(x2+4)+1x2+4 2x2−1x2(x2+4)=941x2+4-141x2∫2x2−1+x2x2+4x2(x2+4) dx =∫941x2+4−141x2+1x2+4dx =98tan-1x2+141x+sin h-1x2+C

∫2x2−1+x2x2+4x2(x2+4) dx=

∫2x2−1+x2x2+4x2(x2+4) dx 2x2−1+x2x2+4x2(x2+4)=2x2−1x2(x2+4)+1x2+4 2x2−1x2(x2+4)=941x2+4-141x2∫2x2−1+x2x2+4x2(x2+4) dx =∫941x2+4−141x2+1x2+4dx =98tan-1x2+141x+sin h-1x2+C

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

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$\frac{9}{8} \tan^{- 1} \frac{x}{2} + \frac{1}{4 x} + \cosh^{- 1} \frac{x}{2} + C$

$\frac{9}{8} \tan^{- 1} \frac{x}{2} + \frac{1}{4 x} + \sin h^{- 1} (\frac{x}{2}) + C$

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

$\frac{9}{16} \log |\frac{2 - x}{2 + x}| + \frac{1}{4 x} + \cosh^{- 1} (\frac{x}{2}) + C$

answer is B.

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Detailed Solution

$\int \frac{2 x^{2} - 1 + x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} d x \frac{2 x^{2} - 1 + x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} = \frac{2 x^{2} - 1}{x^{2} (x^{2} + 4)} + \frac{1}{\sqrt{x^{2} + 4}} \frac{2 x^{2} - 1}{x^{2} (x^{2} + 4)} = \frac{9}{4} \frac{1}{x^{2} + 4} - \frac{1}{4} \frac{1}{x^{2}}$
$\int \frac{2 x^{2} - 1 + x^{2} \sqrt{x^{2} + 4}}{x^{2} (x^{2} + 4)} d x = \int \frac{9}{4} \frac{1}{x^{2} + 4} - \frac{1}{4} \frac{1}{x^{2}} + \frac{1}{\sqrt{x^{2} + 4}} d x = \frac{9}{8} \tan^{- 1} \frac{x}{2} + \frac{1}{4} \frac{1}{x} + \sin h^{- 1} (\frac{x}{2}) + C$