The value of the integral ∫ 0 1 x d x ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) is:

Step-by-step solution

Let x = t ⇒ x = t 2 ,   d x = 2 t d t Let I = ∫ 0 1 x d x ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) I = ∫ 0 1   2 t 2 d t t 2 + 1 3 t 2 + 1 t 2 + 3 = ∫ 0 1   3 t 2 + 1 − t 2 + 1 d t t 2 + 1 3 t 2 + 1 t 2 + 3 = ∫ 0 1   1 t 2 + 3 t 2 + 1 − 1 t 2 + 3 3 t 2 + 1 d t = ∫ 0 1   d t 2 t 2 + 1 − 1 8 3 d t 3 t 2 + 1 − 3 8 d t t 2 + 3 = 1 2 tan − 1 ⁡ t − 3 3 8 × 3 tan − 1 ⁡ 3 t − 3 8 3 tan − 1 ⁡