Definite Integration
Rational Functions with Arctangent Results
Grade 12
Question:
<p>\(\int_2^3 \frac{2x^2}{x^4 + 3x^2 + 1} dx\) is equal to</p>
<p>(a) \(\tan^{-1}\left(\frac{7}{54}\right) + \tan^{-1}\left(\frac{5}{56}\right)\)</p>
<p>(b) \(\tan^{-1}\left(\frac{5}{54}\right) + \tan^{-1}\left(\frac{5}{26}\right)\)</p>
<p>(c) \(\frac{1}{5}\tan^{-1}\left(\frac{7}{54}\right) + \tan^{-1}\left(\frac{5}{56}\right)\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Decompose the integrand strategically and recognize standard arctangent integral forms after substitution or partial fractions.
<p><strong>Solution:</strong> Let $I = \int_2^3 \frac{2x^2}{x^4 + 3x^2 + 1} dx$</p><p>We can decompose:</p><p>$I = \int_2^3 \frac{x^2 + 1}{x^4 + 3x^2 + 1} dx + \int_2^3 \frac{x^2 - 1}{x^4 + 3x^2 + 1} dx$</p><p>The first integral can be written as $\int_2^3 \frac{1}{x^2 + 3 + \frac{1}{x^2}} dx$ and the second as $\int_2^3 \frac{1}{x^2 + 1 + \frac{1}{x^2}} dx$</p><p>After simplification and evaluation at the limits, we get:</p><p>$I = \frac{1}{5}\tan^{-1}\left(\frac{7}{54}\right) + \tan^{-1}\left(\frac{5}{56}\right)$</p><p>Therefore, the answer is (c).</p>
Correct Answer: c