Evaluate:\(\int{\sqrt{x^2-16}}dx\)∫√{x2 - 16} dx

1.	Evaluate:\(\int{\sqrt{x^2-16}}dx\)∫√{x2 - 16} dx
Answer» We know that, \(\int{\sqrt{x^2-a^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-a^2}}\) - \(\frac{a^2}2\) log \| x + \({\sqrt{x^2-a^2}}\) + c \(\int{\sqrt{x^2-4^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-16}}\) - 8 log \| x + \({\sqrt{x^2-16}}\) + c

Answer»

We know that,

\(\int{\sqrt{x^2-a^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-a^2}}\) - \(\frac{a^2}2\) log | x + \({\sqrt{x^2-a^2}}\) + c

\(\int{\sqrt{x^2-4^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-16}}\) - 8 log | x + \({\sqrt{x^2-16}}\) + c

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