Evaluate the integral: int frac{x^4(x^{10}-1) | vedclass

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Question

(B) Let $I = \int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$. Divide numerator and denominator by $x^{10}$:$I = \int \frac{x^4(x^5 - x^{-5})}{x^{10} +

Vedclass · Accepted Answer

$\frac{1}{5} 	an^{-1} (x^5 + \frac{1}{x^5}) + c$

Vedclass · Answer

(B) Let $I = \int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$. Divide numerator and denominator by $x^{10}$:$I = \int \frac{x^4(x^5 - x^{-5})}{x^{10} + 3 + x^{-10}} dx = \int \frac{x^9 - x^{-1}}{x^{10} + 3 + x^{-10}} dx$. This approach is complex. Instead, divide by $x^{10}$ inside the integral differently:$I = \int \frac{x^{14}(1 - x^{-10})}{x^{20}(1 + 3x^{-10} + x^{-20})} dx = \int \frac{x^{-6}(1 - x^{-10})}{1 + 3x^{-10} + x^{-20}} dx$.Let $u = x^5 + x^{-5}$. Then $du = (5x^4 - 5x^{-6}) dx = 5x^4(1 - x^{-10}) dx$.$I = \frac{1}{5} \int \frac{du}{u^2 + 1} = \frac{1}{5} 	an^{-1}(u) + c = \frac{1}{5} 	an^{-1}(x^5 + x^{-5}) + c$.

Evaluate the integral: $\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$

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