If the integral I = ∫ x 5 1 + x 3 d x = K x 3 + 1 x 3 - 2 + C , (where, C is the constant of integration), then the value of 9 K is equal to

Question

If the integral I = ∫ x 5 1 + x 3 d x = K x 3 + 1 x 3 - 2 + C , (where, C is the constant of integration), then the value of 9 K is equal to, 4, 2, 6, 10

MARKS App · Accepted Answer

Let 1 + x 3 = t 2 ⇒ 3 x 2 d x = 2 t d t ⇒ I = ∫ t 2 - 1 2 t d t 3 t ⇒ I = 2 3 ∫ t 2 - 1 d t = 2 t 3 9 - 2 3 t + C = 2 9 t t 2 - 3 + C = 2 9 1 + x 3 x 3 - 2 + C ∴ K = 2 9 ⇒ 9 K = 2

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