Definite Integration
Grade None
Question:
<p>Given that I(n) = <span class="math-tex">\(\int \limits_{1}^{e}\)</span>(1 + log x)<sup>n</sup> dx, n <span class="math-tex">\(\in\)</span> N satisfies I(n) = 2<sup>n</sup> e - 1 - nI(n - 1).<br />
The value of <span class="math-tex">\(\int \limits_{1}^{e}\)</span>(5 + log x)(1 + log x)<sup>2</sup> dx is</p>
<p style="display:inline">5e - 2</p>
<p style="display:inline">10e + 2</p>
<p style="display:inline">5e + 2</p>
<p style="display:inline">10e - 2</p>
Step-by-Step Solution
Key Concept: Decompose the integrand into a linear combination of terms in the form of the given sequence to apply the recurrence relation systematically.
<p>Let I = <span class="math-tex">$\int \limits_{1}^{e}$</span>(5 + log x) (1 + log x)<sup>2</sup> dx<br />
= <span class="math-tex">$\int \limits_{1}^{e}$</span>(4 + 1 + log x)(1 + log x)<sup>2</sup>dx<br />
= 4<span class="math-tex">$\int \limits_{1}^{e}$</span>(1 + log x)<sup>2</sup>dx + <span class="math-tex">$\int \limits_{1}^{e}$</span>(1 + log x)<sup>3</sup>dx<br />
= 4[4e - 1 - 2I (1)] + [8e - 1 - 3I(2)] ...(i)<br />
I(1) = 2e - 1 - 1 (0) = 2e - 1 - (e - 1) = e ...(ii)<br />
I(2) = 4e - 1 - 2I(1)<br />
<span class="math-tex">$\Leftrightarrow$</span> I(2) = 4e - 1 - 2e = 2e - 1 ...(iii)<br />
From (i), (ii) and (iii), we get<br />
I = 4 (4e - 1 - 2e) + 8e - 1 - 3(2e - 1)<br />
= 10e - 2</p>
Correct Answer: D