MHT CET202620 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
(x²¹ + x^6 + x^3)(2x¹⁸ + 7x^3 + 14)^ 1 3 dx =
Options
- A1 56 (2x¹⁸ + 7x^3 + 14)^ 4 3 + c
- B(2x¹⁸ + 7x^3 + 14)^ 4 3 + c
- C(2x²¹ + 7x^6 + 14x^3)^ 4 3 + c
- D1 56 (2x²¹ + 7x^6 + 14x^3)^ 4 3 + c
Correct answer
D. 1 56 (2x²¹ + 7x^6 + 14x^3)^ 4 3 + c
Step-by-step solution
Let the given integral be I = (x²¹ + x^6 + x^3)(2x¹⁸ + 7x^3 + 14)^ 1 3 dx . Factoring out x from the first polynomial, we get: I = x(x²⁰ + x^5 + x^2)(2x¹⁸ + 7x^3 + 14)^ 1 3 dx Moving x inside the cube root as x^3 , we obtain: I = (x²⁰ + x^5 + x^2)(2x²¹ + 7x^6 + 14x^3)^ 1 3 dx Let 2x²¹ + 7x^6 + 14x^3 = t . Differentiating both sides with respect to x : (42x²⁰ + 42x^5 + 42x^2) dx = dt 42(x²⁰ + x^5 + x^2) dx = dt (x²⁰ + x^5 + x^2) dx = dt 42 Substituting this into the integral, we get: I = t^ 1 3 dt 42 I = 1 42 t^ 4 3