$\int (6 x^{2} + 5 x + 4) (x^{2} + x + 1)^{6} . x^{27} d x$ equals (where $c$ is integration constant)

\frac{x^{28} (1 + x + x^{2})^{7}}{7} + c

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\frac{x^{24} (1 + x + x^{2})^{7}}{7} + c

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\frac{x^{28} (1 + x + x^{2})^{8}}{7} + c

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\frac{x^{24} (1 + x + x^{2})^{6}}{7} + c

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Solution

The correct option is A $\frac{x^{28} (1 + x + x^{2})^{7}}{7} + c$
$\int (x^{2} + x + 1)^{6} . x^{24} (6 x^{5} + 5 x^{4} + 4 x^{3}) d x$
$= \int (x^{6} + x^{5} + x^{4})^{6} (6 x^{5} + 5 x^{4} + 4 x^{3}) d x$
$L e t x^{6} + x^{5} + x^{4} = t$
$(6 x^{5} + 5 x^{4} + 4 x^{3}) d x = d t$
$= \int t^{6} d t = \frac{t^{7}}{7} + c = \frac{x^{28} (x^{2} + x + 1)^{7}}{7}$

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