If int frac{2 x^{12}+5 x^9}{left(1+x^3+x^5rig | vedclass

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(B) Let $I = \int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x$. Divide the numerator and denominator by $x^{15}$ inside the integral: $I = \in

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $I = \int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x$. Divide the numerator and denominator by $x^{15}$ inside the integral: $I = \int \frac{2 x^{12}+5 x^9}{x^{15} \left(\frac{1}{x^5}+\frac{x^3}{x^5}+\frac{x^5}{x^5}\right)^3} d x = \int \frac{2 x^{-3}+5 x^{-6}}{\left(x^{-5}+x^{-2}+1\right)^3} d x$. Let $t = 1 + x^{-2} + x^{-5}$. Then $dt = (-2x^{-3} - 5x^{-6}) dx$, which implies $-(2x^{-3} + 5x^{-6}) dx = dt$. Substituting this into the integral: $I = -\int \frac{dt}{t^3} = -\int t^{-3} dt = -\frac{t^{-2}}{-2} + C = \frac{1}{2t^2} + C$. Substituting $t$ back: $I = \frac{1}{2 \left(1 + \frac{1}{x^2} + \frac{1}{x^5}\right)^2} + C = \frac{1}{2 \left(\frac{x^5+x^3+1}{x^5}\right)^2} + C = \frac{x^{10}}{2(1+x^3+x^5)^2} + C$. Comparing this with $\frac{x^m}{l(1+x^3+x^5)^r} + C$, we get $m=10$, $l=2$, and $r=2$. Therefore, $\frac{m-l}{r} = \frac{10-2}{2} = \frac{8}{2} = 4$.

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$ then $\frac{m-l}{r}=$

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