Indefinite Integration
Integration of the form e^x[f(x)+f'(x)]
Grade 12
Question:
<p>The integral \(\int\left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\) is equal to</p>
<p>\((x+1)e^{x+\frac{1}{x}}+C\)</p>
<p>\(-xe^{x+\frac{1}{x}}+C\)</p>
<p>\((x-1)e^{x+\frac{1}{x}}+C\)</p>
<p>\(xe^{x+\frac{1}{x}}+C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be written as the derivative of e^(x+1/x) by identifying that d/dx[e^(x+1/x)] = e^(x+1/x)·(1-1/x²). Rewrite the integrand to match this form using algebraic manipulation.
<p><strong>Step 1:</strong> Find the derivative of e^(x+1/x) to guide our approach.</p><p>Let u = x + 1/x, then du/dx = 1 - 1/x²</p><p>d/dx[e^(x+1/x)] = e^(x+1/x)·(1 - 1/x²)</p><p><strong>Step 2:</strong> Rewrite the integrand to match this pattern.</p><p>1 + x - 1/x = (1 - 1/x²)·x + 1 = (1 - 1/x²)·x + (1 - 1/x²) + 1/x²</p><p>Actually, observe: (1 + x - 1/x)·e^(x+1/x) = d/dx[x·e^(x+1/x)]</p><p>Verify: d/dx[x·e^(x+1/x)] = e^(x+1/x) + x·e^(x+1/x)·(1 - 1/x²) = e^(x+1/x) + e^(x+1/x)·(x - 1/x) = (1 + x - 1/x)·e^(x+1/x) ✓</p><p><strong>Step 3:</strong> Integrate directly.</p><p>∫(1 + x - 1/x)·e^(x+1/x)dx = x·e^(x+1/x) + C</p><p>∴ Answer: <strong>x·e^(x+1/x) + C</strong></p>
Correct Answer: D