int {left( {6{x^2} + 5x + 4} right){{left( {{ | vedclass

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(A) Given integral: $I = \int (6x^2 + 5x + 4)(x^2 + x + 1)^6 \cdot x^{27} dx$We can rewrite the expression as:$I = \int (x^2 + x + 1)^6 \cdot x^{24} \

Vedclass · Accepted Answer

$\frac{{{x^{28}}{{\left( {1 + x + {x^2}} \right)}^7}}}{7} + C$

Vedclass · Answer

(A) Given integral: $I = \int (6x^2 + 5x + 4)(x^2 + x + 1)^6 \cdot x^{27} dx$We can rewrite the expression as:$I = \int (x^2 + x + 1)^6 \cdot x^{24} \cdot (6x^2 + 5x + 4)x^3 dx$$I = \int (x^2 + x + 1)^6 \cdot x^{24} \cdot (6x^5 + 5x^4 + 4x^3) dx$$I = \int (x^2(x^2 + x + 1))^6 \cdot (6x^5 + 5x^4 + 4x^3) dx$$I = \int (x^6 + x^5 + x^4)^6 \cdot (6x^5 + 5x^4 + 4x^3) dx$Let $t = x^6 + x^5 + x^4$.Then $dt = (6x^5 + 5x^4 + 4x^3) dx$.Substituting these into the integral:$I = \int t^6 dt = \frac{t^7}{7} + C$$I = \frac{(x^6 + x^5 + x^4)^7}{7} + C$$I = \frac{(x^4(x^2 + x + 1))^7}{7} + C$$I = \frac{x^{28}(x^2 + x + 1)^7}{7} + C$

$\int {\left( {6{x^2} + 5x + 4} \right){{\left( {{x^2} + x + 1} \right)}^6} \cdot {x^{27}}dx} $ equals (where $C$ is integration constant)

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