The value of 2 -1 ((loge x) +1) 4 e \int 2 e 1 x ( 2 e -1 2 -1 ) dx is ((loge x) +1) ((6-loge x) +1) e +e
Step-by-Step Solution
Key Concept: Apply the core result for definite integral properties and simplify using the given constraints.
x t 1 (3) Put ln x = t \Rightarrow dx = dt e 2 2 x 4 e 4 -1 4 2 (t +1) e I = \int dt \ldots (i) 2 -1 -1 2 2 (t +1) ((6-t) +1) e + e 2 -1 4 ((6-t) +1) e I = \int \ldots (ii) -1 -1 2 2 2 ((6-t) +1) (t +1) e + e dt 0 0 { Using \int f (x)dx = \int f (a + b - x)dx} a a Adding (i) and (ii) gives 2l = \int dt \Rightarrow l = 1 3
Correct Answer: 3