Indefinite Integration
Integration using Substitution
Grade 12
Question:
<p>\(\int \frac{x^2}{x^4 + x^2 + 1} dx\) is equal to</p>
<p>(a) \(\log(x^4 + x^2 + 1) + C\)</p>
<p>(b) \(\log\frac{x^2}{x^4 - x^2 + 1} + C\)</p>
<p>(c) \(\frac{1}{\sqrt{3}}\tan^{-1}\frac{2x^2 - 1}{\sqrt{3}} + C\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Divide by highest power in denominator, then use substitution $u = x - \frac{1}{x}$ to convert to standard arctangent form.
<p>Divide numerator and denominator by $x^2$: $\int \frac{1}{x^2 + 1 + \frac{1}{x^2}} dx$</p><p>Let $u = x - \frac{1}{x}$, then $du = \left(1 + \frac{1}{x^2}\right) dx$</p><p>$x^2 + \frac{1}{x^2} = u^2 + 2$, so the integral becomes $\int \frac{du}{u^2 + 3} = \frac{1}{\sqrt{3}}\tan^{-1}\frac{u}{\sqrt{3}} + C = \frac{1}{\sqrt{3}}\tan^{-1}\frac{2x^2 - 1}{\sqrt{3}} + C$</p>
Correct Answer: C