Evaluate ∫ (x^2 + 1)/(x^2 - 1) dx.

Question 68

HardOriginal question

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

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

Correct answerYour answer

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

Correct answerYour answer

x + 2\log\left|x^2 - 1\right| + C

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

The fraction is improper (equal degrees), so divide first: $\frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}$ .
$\frac{2}{x^2 - 1} = \frac{1}{x - 1} - \frac{1}{x + 1}$ , so the integral is $x + \log\left|\frac{x - 1}{x + 1}\right| + C$ .

Worth noting

Leaving out the $x$ forgets the 1 that the division gives.

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