Q.2 (B)(i) Evaluate : \(\displaystyle\int \frac{x}{4x^4 - 20x^2 - 3}\,dx\)
Best answer
Answer: \(\dfrac{1}{16\sqrt{7}}\log\left|\dfrac{2x^2 - 5 - 2\sqrt{7}}{2x^2 - 5 + 2\sqrt{7}}\right| + c\)
Put \(x^2 = t\), so \(2x\,dx = dt\), i.e. \(x\,dx = \dfrac{dt}{2}\).
\(\displaystyle I = \frac{1}{2}\int \frac{dt}{4t^2 - 20t - 3} = \frac{1}{8}\int \frac{dt}{t^2 - 5t - \frac{3}{4}}\)
Completing the square: \(t^2 - 5t - \dfrac{3}{4} = \left(t - \dfrac{5}{2}\right)^2 - \dfrac{25}{4} - \dfrac{3}{4} = \left(t - \dfrac{5}{2}\right)^2 - (\sqrt{7})^2\).
\(\displaystyle I = \frac{1}{8}\int \frac{dt}{\left(t - \frac{5}{2}\right)^2 - (\sqrt{7})^2}\)
Using \(\displaystyle\int \frac{dX}{X^2 - a^2} = \frac{1}{2a}\log\left|\frac{X - a}{X + a}\right| + c\):
\(\displaystyle I = \frac{1}{8} \cdot \frac{1}{2\sqrt{7}}\log\left|\frac{t - \frac{5}{2} - \sqrt{7}}{t - \frac{5}{2} + \sqrt{7}}\right| + c = \frac{1}{16\sqrt{7}}\log\left|\frac{2x^2 - 5 - 2\sqrt{7}}{2x^2 - 5 + 2\sqrt{7}}\right| + c\)
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