Evaluate ∫ 1/(√(x^2 - 16)) dx.

Question 52

MediumOriginal question

Evaluate $\int \frac{1}{\sqrt{x^{2} - 16}}\,dx$ .

\log\left|x + \sqrt{x^{2} + 16}\right| + C

Correct answerYour answer

\log\left|x + \sqrt{x^{2} - 16}\right| + C

Correct answerYour answer

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

Correct answerYour answer

\sqrt{x^{2} - 16} + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$\int \frac{dx}{\sqrt{x^{2} - a^{2}}} = \log\left|x + \sqrt{x^{2} - a^{2}}\right| + C$ .
With $a = 4$ : $\log\left|x + \sqrt{x^{2} - 16}\right| + C$ .

Worth noting

This formula has no $\frac{1}{a}$ in front, unlike $\int \frac{dx}{x^{2} + a^{2}}$ .

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