Indefinite Integration
Integration by Parts / Substitution
Grade 12
Question:
<p>The evaluation of \(\displaystyle\int e^{x-\frac{1}{x}}\cdot\frac{x^2+1}{x^2}\,dx\) is</p>
<li>\(e^{x-\frac{1}{x}}+C\)</li>
<li>\(e^{x+\frac{1}{x}}+C\)</li>
<li>\(x\,e^{x-\frac{1}{x}}+C\)</li>
<li>\(\dfrac{1}{x}\,e^{x+\frac{1}{x}}+C\)</li>
Step-by-Step Solution
Key Concept: Recognise the form \inteᵍ⁽ˣ⁾ \cdot g'(x)dx = eᵍ⁽ˣ⁾+C with g(x)=x-1/x, g'(x)=1+1/x^2.
<p><strong>Key recognition:</strong> Let $g(x)=x-\dfrac{1}{x}$. Then $g'(x)=1+\dfrac{1}{x^2}=\dfrac{x^2+1}{x^2}$.</p>
<p>The integrand is exactly $e^{g(x)}\cdot g'(x)$, so:</p>
<p>$$\int e^{x-\frac1x}\cdot\frac{x^2+1}{x^2}\,dx = e^{x-\frac1x}+C$$</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A