int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14) | vedclass

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Question

(A) Let $I = \int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{1/3} dx$. Factor out $x^{18}$ from the first term: $I = \int x^{18}(x^3 + x^{-12} + x^{-1

Vedclass · Accepted Answer

$\frac{1}{56} (2x^{18} + 7x^3 + 14)^{4/3} + c$

Vedclass · Answer

(A) Let $I = \int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{1/3} dx$. Factor out $x^{18}$ from the first term: $I = \int x^{18}(x^3 + x^{-12} + x^{-15})(2x^{18} + 7x^3 + 14)^{1/3} dx$. Rewrite the integral as $I = \int x^{18} \cdot x^{18/3} (2 + 7x^{-15} + 14x^{-18})^{1/3} (x^3 + x^{-12} + x^{-15}) dx$. $I = \int x^{24} (2 + 7x^{-15} + 14x^{-18})^{1/3} (x^3 + x^{-12} + x^{-15}) dx$. Let $u = 2 + 7x^{-15} + 14x^{-18}$. Then $du = (7(-15)x^{-16} + 14(-18)x^{-19}) dx = (-105x^{-16} - 252x^{-19}) dx = -21x^{-19}(5x^3 + 12) dx$. Alternatively, factor out $x^{18}$ from the bracket: $I = \int x^{18} (x^3 + x^{-12} + x^{-15}) (x^{18})^{1/3} (2 + 7x^{-15} + 14x^{-18})^{1/3} dx = \int x^{24} (x^3 + x^{-12} + x^{-15}) (2 + 7x^{-15} + 14x^{-18})^{1/3} dx$. Let $t = 2x^{18} + 7x^3 + 14$. Then $dt = (36x^{17} + 21x^2) dx = 3x^2(12x^{15} + 7) dx$. $I = \int x^2(x^{19} + x^4 + x) (2x^{18} + 7x^3 + 14)^{1/3} dx$. $I = \frac{1}{56} (2x^{18} + 7x^3 + 14)^{4/3} + c$.

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

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