Question
Evaluate the following:
$\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Hint: Put $\text{x}^2=\sec\theta$
$\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Hint: Put $\text{x}^2=\sec\theta$
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$\text{x}_\text{i}$
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$0$
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$1$
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$2$
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$3$
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$4$
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$5$
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$\text{p}_\text{i}$
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$\frac{1}{6}$
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$\frac{5}{18}$
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$\frac{2}{9}$
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$\frac{1}{6}$
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$\frac{1}{9}$
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$\frac{1}{18}$
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