int frac{x^5 , dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x | vedclass

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Question

(B) Let the integral be $I = \int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)}$.First,simplify the denominator terms:$(x^2+x+1)(x^4-x^3+x-1) = (x^2

Vedclass · Accepted Answer

$\frac{1}{12} \log_e \left| \frac{x^6-1}{x^6+1} \right| + c$

Vedclass · Answer

(B) Let the integral be $I = \int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)}$.First,simplify the denominator terms:$(x^2+x+1)(x^4-x^3+x-1) = (x^2+x+1)(x^3(x-1) + 1(x-1)) = (x^2+x+1)(x^3+1)(x-1)$.Since $(x^2+x+1)(x-1) = x^3-1$,the product becomes $(x^3-1)(x^3+1) = x^6-1$.Thus,the integral simplifies to $I = \int \frac{x^5 \, dx}{(x^6+1)(x^6-1)}$.Substitute $t = x^6$,then $dt = 6x^5 \, dx$,so $x^5 \, dx = \frac{1}{6} \, dt$.$I = \frac{1}{6} \int \frac{dt}{(t+1)(t-1)}$.Using partial fractions,$\frac{1}{(t+1)(t-1)} = \frac{1}{2} \left( \frac{1}{t-1} - \frac{1}{t+1} \right)$.$I = \frac{1}{6} \cdot \frac{1}{2} \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) dt = \frac{1}{12} (\log_e |t-1| - \log_e |t+1|) + c$.$I = \frac{1}{12} \log_e \left| \frac{t-1}{t+1} \right| + c = \frac{1}{12} \log_e \left| \frac{x^6-1}{x^6+1} \right| + c$.

$\int \frac{x^5 \, dx}{(x^2+x+1)(x^6+1)(x^4-x^3+x-1)} =$

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