Q88 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics

$\int(x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{\frac{1}{3}}\ dx =$

  • A
    $\dfrac{1}{56}(2x^{18} + 7x^3 + 14)^{\frac{4}{3}} + c$
  • B
    $(2x^{18} + 7x^3 + 14)^{\frac{4}{3}} + c$
  • C
    $(2x^{21} + 7x^6 + 14x^3)^{\frac{4}{3}} + c$
  • Correct answer
    D
    $\dfrac{1}{56}(2x^{21} + 7x^6 + 14x^3)^{\frac{4}{3}} + c$

Solution

Let the given integral be $I = \int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{\dfrac{1}{3}}\ dx$.

Factoring out $x$ from the first polynomial, we get:

$I = \int x(x^{20} + x^5 + x^2)(2x^{18} + 7x^3 + 14)^{\dfrac{1}{3}}\ dx$

Moving $x$ inside the cube root as $x^3$, we obtain:

$I = \int (x^{20} + x^5 + x^2)(2x^{21} + 7x^6 + 14x^3)^{\dfrac{1}{3}}\ dx$

Let $2x^{21} + 7x^6 + 14x^3 = t$.

Differentiating both sides with respect to $x$:

$(42x^{20} + 42x^5 + 42x^2)\ dx = dt$

$42(x^{20} + x^5 + x^2)\ dx = dt$

$(x^{20} + x^5 + x^2)\ dx = \dfrac{dt}{42}$

Substituting this into the integral, we get:

$I = \int t^{\dfrac{1}{3}} \dfrac{dt}{42}$

$I = \dfrac{1}{42} \dfrac{t^{\dfrac{4}{3}}}{\dfrac{4}{3}} + c$

$I = \dfrac{1}{56} t^{\dfrac{4}{3}} + c$

Substituting back $t = 2x^{21} + 7x^6 + 14x^3$:

$I = \dfrac{1}{56} (2x^{21} + 7x^6 + 14x^3)^{\dfrac{4}{3}} + c$

Hence, the correct answer is option: 'D'.

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