The integral ∫ 1 e x e 2 x - e x x l o g e   x    d x is equal to

Step-by-step solution

Let I 1 = ∫ 1 e x e 2 x log e   x   ⁡ d x Let x e x = t Taking natural logarithm on both sides, we get x   ln x e = ln  t ⇒ x ln  x − 1 = ln   t     . . . . . . . . . i x t 1 1 ln 1 - 1 = ln t ⇒ t = 1 e e e ln e - 1 = ln t ⇒ t = 1 Differentiating equation i on both the sides, we get ln   x − 1 + x 1 x d x = 1 t d t ⇒ln   x   d x = 1 t d t ∴ I 1 = ∫ 1 e 1 t 2 1 t d t = t 2 2 1 e 1 = 1 2 - 1 2 e 2