≤ft(x²1}+x^{6}+x³)≤ft(2 x¹8}+7 x³+14)¹/3} d x=

Correct Answer is : 156(2x21+7x6+14x3)43+c1/56≤ft(2 x²1}+7 x^{6}+14 x³)^{4/3}+c561(2x21+7x6+14x3)34+c

1/56≤ft(2 x¹8}+7 x³+14)^{4/3}+c

≤ft(2 x¹8}+7 x³+14)^{4/3}+c

≤ft(2 x²1}+7 x^{6}+14 x³)^{4/3}+c

1/56≤ft(2 x²1}+7 x^{6}+14 x³)^{4/3}+c

Correct Answer is : 156(2x21+7x6+14x3)43+c1/56≤ft(2 x²1}+7 x^{6}+14 x³)^{4/3}+c561(2x21+7x6+14x3)34+c

Test Index

MHTCET 20 Apr 2026 Shift 2 Paper

Question : 131 of 150

Marks: +1, -0

\int\left(x^{21}+x^{6}+x^{3}\right)\left(2 x^{18}+7 x^{3}+14\right)^{\frac{1}{3}} d x=

Solution:

Concept:

Use substitution so the derivative matches the given integrand.

Explanation:

Let

u = 2x^{21} + 7x^{6} + 14x^{3}

Differentiate with respect to

x

\frac{du}{dx} = 42x^{20} + 42x^{5} + 42x^{2}

Factor the derivative:

du = 42x^{2}(x^{18} + x^{3} + 1)\,dx

Rewrite the integrand using

x^{21}+x^{6}+x^{3} = x^{3}(x^{18}+x^{3}+1)

x^{3}(x^{18}+x^{3}+1)(2x^{18}+7x^{3}+14)^{\frac{1}{3}}\,dx

Since

u^{\frac{1}{3}} = x(2x^{18}+7x^{3}+14)^{\frac{1}{3}}

, the integrand becomes:

x^{2}(x^{18}+x^{3}+1)u^{\frac{1}{3}}\,dx

Using

du = 42x^{2}(x^{18}+x^{3}+1)\,dx

, the integral becomes:

\frac{1}{42}\int u^{\frac{1}{3}}\,du

Evaluate:

\frac{1}{42} \cdot \frac{u^{\frac{4}{3}}}{\frac{4}{3}} + c = \frac{1}{56}u^{\frac{4}{3}} + c

Substitute back

u = 2x^{21} + 7x^{6} + 14x^{3}

\frac{1}{56}(2x^{21}+7x^{6}+14x^{3})^{\frac{4}{3}} + c

Answer:

\frac{1}{56}\left(2x^{21}+7x^{6}+14x^{3}\right)^{\frac{4}{3}} + c

Hence, Option D.

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