Indefinite Integration
Substitution Method
Grade 12
Question:
<p>\(\int \frac{x^2 - 1}{(x^4 + 3x^2 + 1)\tan^{-1}\left(x + \frac{1}{x}\right)} dx\) is equal to</p>
<p>(a) \(\tan^{-1}\left(x + \frac{1}{x}\right) + C\)</p>
<p>(b) \(\cot^{-1}\left(x + \frac{1}{x}\right) + C\)</p>
<p>(c) \(\log \tan^{-1}\left(x + \frac{1}{x}\right) + C\)</p>
<p>(d) \(\log\left|x + \frac{1}{x}\right| + C\)</p>
Step-by-Step Solution
Key Concept: Recognize the derivative of the inverse tangent function and use substitution to convert to a standard logarithmic integral.
<p><strong>Solution:</strong> Let $u = \tan^{-1}\left(x + \frac{1}{x}\right)$. Then $du = \frac{1}{1+\left(x+\frac{1}{x}\right)^2} \cdot \frac{x^2-1}{x^2} dx = \frac{x^2-1}{x^4+3x^2+1} dx$. The integral becomes $\int \frac{du}{u} = \log|u| + C = \log\tan^{-1}\left(x + \frac{1}{x}\right) + C$.</p>
Correct Answer: C