If m is any natural number, then the value of the integral ∫ x 3 m + x 2 m + x m 2 x 2 m + 3 x m + 6 1 / m dx is (where, C is an arbitrary constant)

Step-by-step solution

Put, I = ∫ x 3 m + x 2 m + x m 2 x 3 m + 3 x 2 m + 6 x m 1 / m x d x = ∫ x 3 m - 1 + x 2 m - 1 + x m - 1 2 x 3 m + 3 x 2 m + 6 x m 1 / m d x = 1 6 m ∫ 2 x 3 m + 3 x 2 m + 6 x m 1 / m 6 mx 3 m - 1 + 6 mx 2 m - 1 + 6 mx m - 1 dx Now, let 2 x 3 m + 3 x 2 m + 6 x m = t ⇒ 6 mx 3 m - 1 + 6 mx 2 m - 1 + 6 mx m - 1 dx = dt ⇒ I = 1 6 m ∫ t 1 / m dt = 1 6 m t 1 / m + 1 1 / m + 1 = 1 6 m + 1 2 x 3 m + 3 x 2 m + 6 x m (1 / m) + 1 + C