<p>If <span>\(I_n = \int_1^e (\ln x)^n dx\)</span>, then <span>\(I_n + nI_{n-1}\)</span> is equal to</p>
Step-by-Step Solution
Key Concept: Apply integration by parts strategically to relate consecutive integrals and establish a recurrence relation.
<p>Using integration by parts: <span>$I_n = \int_1^e (\ln x)^n dx$</span>. Let <span>$u = (\ln x)^n$</span> and <span>$dv = dx$</span>. Then <span>$du = n(\ln x)^{n-1}\frac{1}{x}dx$</span> and <span>$v = x$</span>.</p><p>$$I_n = [x(\ln x)^n]_1^e - \int_1^e x \cdot n(\ln x)^{n-1}\frac{1}{x}dx = e - nI_{n-1}$$</p><p>Therefore <span>$I_n + nI_{n-1} = e$</span>.</p>
Correct Answer: B