Consider Ι α = ∫ α α 2 d x x (where α > 0 ), then the value of ∑ r = 2 5 Ι r + ∑ k = 2 5 Ι 1 k is

Step-by-step solution

Ι α = ln ⁡ x α α 2 = ln ⁡ α 2 - ln ⁡ α = ln ⁡ α 2 α = ln ⁡ α Thus, ∑ r = 2 5 Ι r = ln ⁡ 2 + ln ⁡ 3 + ln ⁡ 4 + ln ⁡ 5 and ∑ r = 2 5 Ι 1 k = ln ⁡ 1 2 + ln ⁡ 1 3 + ln ⁡ 1 4 + ln ⁡ 1 5 = - ln ⁡ 2 + ln ⁡ 3 + ln ⁡ 4 + ln ⁡ 5 Hence, their sum equals to zero