If int limits_0^1 (x^{21}+x^{14}+x^7)(2x^{14} | vedclass

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(C) Let $I = \int \limits_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx$.We can rewrite the integrand by factoring $x^7$ from the second term: $I =

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$63$

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(C) Let $I = \int \limits_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx$.We can rewrite the integrand by factoring $x^7$ from the second term: $I = \int \limits_0^1 (x^{21}+x^{14}+x^7) \cdot x \cdot (2x^7+3+6x^{-7})^{1/7} dx$ is not the most efficient way. Instead,let $t = 2x^{21} + 3x^{14} + 6x^7$. Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42(x^{20} + x^{13} + x^6) dx$.However,the original integral is $\int_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx$. Let $u = 2x^{14} + 3x^7 + 6$. Then $du = (28x^{13} + 21x^6) dx = 7(4x^{13} + 3x^6) dx$. This does not match the first factor.Re-evaluating: The integral is $\int_0^1 x^7(x^{14}+x^7+1)(2x^{14}+3x^7+6)^{1/7} dx$. Let $t = 2x^{14} + 3x^7 + 6$. Then $dt = (28x^{13} + 21x^6) dx = 7(4x^{13} + 3x^6) dx$. Actually,the correct substitution is $t = 2x^{21} + 3x^{14} + 6x^7$. Then $dt = (42x^{20} + 42x^{13} + 42x^6) dx = 42x^6(x^{14} + x^7 + 1) dx$.Thus,$\int_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx = \frac{1}{42} \int_0^{11} t^{1/7} dt = \frac{1}{42} [\frac{7}{8} t^{8/7}]_0^{11} = \frac{1}{48} (11)^{8/7}$.Here $l = 48, m = 8, n = 7$. Since $m, n$ are coprime,$l+m+n = 48+8+7 = 63$.

If $\int \limits_0^1 (x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7} dx = \frac{1}{l}(11)^{m/n}$ where $l, m, n \in N$,$m$ and $n$ are coprime,then $l+m+n$ is equal to $...........$.

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