Evaluate ∫ 1/(x^2 - 9) dx.

Question 48

MediumOriginal question

Evaluate $\int \frac{1}{x^{2} - 9}\,dx$ .

\frac{1}{6}\log\left|\frac{x + 3}{x - 3}\right| + C

Correct answerYour answer

\frac{1}{3}\tan^{-1}\frac{x}{3} + C

Correct answerYour answer

\log\left|x^2 - 9\right| + C

Correct answerYour answer

\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$\frac{1}{x^2 - 9} = \frac{1}{6}\left(\frac{1}{x - 3} - \frac{1}{x + 3}\right)$ .
So the integral is $\frac{1}{6}\log\left|\frac{x - 3}{x + 3}\right| + C$ : the standard form $\frac{1}{2a}\log\left|\frac{x - a}{x + a}\right|$ .

Worth noting

The $\tan^{-1}$ form belongs to $x^2 + a^2$ , with a plus sign.

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