Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>\(\displaystyle\int \frac{x}{(x^2+1)(x^2+4)}\,dx\) equals</p>
<li>\(\dfrac{1}{6}\ln\!\left|\dfrac{x^2+1}{x^2+4}\right|+C\)</li>
<li>\(\dfrac{1}{3}\ln\!\left|\dfrac{x^2+4}{x^2+1}\right|+C\)</li>
<li>\(\dfrac{1}{6}\tan^{-1}(x^2)+C\)</li>
<li>\(\dfrac{1}{3}\ln\!\left|\dfrac{x^2+1}{x^2+4}\right|+C\)</li>
Step-by-Step Solution
Key Concept: Substitute t=x^2, dt=2x dx. Then partial fractions: 1/((t+1)(t+4)) = (1/3)[1/(t+1)-1/(t+4)].
<p><strong>Substitution:</strong> Let $t=x^2\Rightarrow dt=2x\,dx$.</p>
<p>$$\frac{1}{2}\int\frac{dt}{(t+1)(t+4)}$$</p>
<p>Partial fractions: $\dfrac{1}{(t+1)(t+4)}=\dfrac{1}{3}\left(\dfrac{1}{t+1}-\dfrac{1}{t+4}\right)$</p>
<p>$$= \frac{1}{6}\ln\!\left|\frac{t+1}{t+4}\right|+C = \frac{1}{6}\ln\!\left|\frac{x^2+1}{x^2+4}\right|+C$$</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A