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(C) The integral $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is a standard integral formula.By using the substitution $x = a 	an 	heta$,we have $dx = a \sec^

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$\log |x + \sqrt{x^2 + a^2}| + c$

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(C) The integral $\int \frac{dx}{\sqrt{x^2 + a^2}}$ is a standard integral formula.By using the substitution $x = a 	an 	heta$,we have $dx = a \sec^2 	heta \ d	heta$.Then,$\sqrt{x^2 + a^2} = \sqrt{a^2 	an^2 	heta + a^2} = a \sec 	heta$.The integral becomes $\int \frac{a \sec^2 	heta \ d	heta}{a \sec 	heta} = \int \sec 	heta \ d	heta$.The integral of $\sec 	heta$ is $\log |\sec 	heta + 	an 	heta| + c$.Substituting back $	an 	heta = \frac{x}{a}$ and $\sec 	heta = \frac{\sqrt{x^2 + a^2}}{a}$,we get $\log |\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}| + c = \log |\frac{x + \sqrt{x^2 + a^2}}{a}| + c = \log |x + \sqrt{x^2 + a^2}| - \log |a| + c$.Since $-\log |a| + c$ is a constant,we write the final result as $\log |x + \sqrt{x^2 + a^2}| + c$.

$\int \frac{dx}{\sqrt{x^2 + a^2}}$ is equal to

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