Indefinite Integration
Integration of Rational Functions
Grade 12
Question:
<p>\(\displaystyle\int \frac{dx}{(x^2-x+1)^2}\) equals (where \(C\) is the constant of integration)</p>
<li>\(\dfrac{2x-1}{3(x^2-x+1)}+\dfrac{2}{3\sqrt3}\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\)</li>
<li>\(\dfrac{2x-1}{2(x^2-x+1)}+\dfrac{1}{\sqrt3}\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\)</li>
<li>\(\dfrac{2x-1}{3(x^2-x+1)}+\dfrac{2\sqrt3}{9}\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\)</li>
<li>\(\dfrac{2x-1}{x^2-x+1}+\dfrac{2}{\sqrt3}\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\)</li>
Step-by-Step Solution
Key Concept: Complete the square: x^2-x+1=(x-½)^2+¾. Use the reduction formula for \intdx/(t^2+a^2)^2 with t=x-½.
<p><strong>Complete the square:</strong> $x^2-x+1 = \left(x-\tfrac12\right)^2+\tfrac34$.</p>
<p>Let $t=x-\tfrac12$, $a^2=\tfrac34$. Use the reduction formula:</p>
<p>$$\int\frac{dt}{(t^2+a^2)^2} = \frac{t}{2a^2(t^2+a^2)}+\frac{1}{2a^3}\tan^{-1}\frac{t}{a}+C$$</p>
<p>With $a^2=3/4$, $a=\sqrt3/2$:</p>
<p>$$= \frac{t}{2\cdot\frac34\cdot(t^2+\frac34)}+\frac{1}{2(\frac{\sqrt3}{2})^3}\tan^{-1}\frac{2t}{\sqrt3}+C$$</p>
<p>$$= \frac{2(2x-1)}{3\cdot3(x^2-x+1)}\cdot\frac32 +\frac{4}{3\sqrt3}\cdot\frac12\tan^{-1}\frac{2x-1}{\sqrt3}+C$$</p>
<p>Simplifying gives option <strong>(C)</strong>.</p>
Correct Answer: C