Question Detail

The given reduction formula is I_n = e²/2 - (n/2)I_(n-1), where I_n = ∫₁^e x(ln x)ⁿ dx. To find I₂, we substitute n=2 into the formula: I₂ = e²/2 - (2/2)I₁ = e²/2 - I₁. Now we need to calculate I₁ = ∫₁^e x(ln x)¹ dx = ∫₁^e x ln x dx. Use integration by parts for I₁: let u = ln x and dv = x dx. Then du = (1/x) dx and v = x²/2. I₁ = [(x²/2) ln x] from 1 to e - ∫₁^e (x²/2)(1/x) dx. Evaluate the first term: [(e²/2)ln e - (1²/2)ln 1] = [ (e²/2)1 - (1/2)0 ] = e²/2. The second term is ∫₁^e (x/2) dx = [x²/4] from 1 to e = (e²/4 - 1²/4) = (e²-1)/4. So, I₁ = e²/2 - (e²-1)/4 = (2e² - e² + 1)/4 = (e² + 1)/4. Finally, substitute I₁ back into the expression for I₂: I₂ = e²/2 - I₁ = e²/2 - (e² + 1)/4 = (2e² - (e² + 1))/4 = (2e² - e² - 1)/4 = (e² - 1)/4.