The value of int frac{2 x^{12}+5 x^9}{left(x^ | vedclass

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(B) To solve the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$, we factor out $x^5$ from the denominator term inside the c

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$\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C$

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(B) To solve the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$, we factor out $x^5$ from the denominator term inside the cube: $I = \int \frac{2 x^{12}+5 x^9}{\left[x^5\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)\right]^3} \,d x$ $I = \int \frac{2 x^{12}+5 x^9}{x^{15}\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$ $I = \int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$ Let $t = 1+\frac{1}{x^2}+\frac{1}{x^5}$. Then $dt = \left(-\frac{2}{x^3}-\frac{5}{x^6}\right) dx$, which implies $-dt = \left(\frac{2}{x^3}+\frac{5}{x^6}\right) dx$. Substituting these 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: $I = \frac{1}{2\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^2} + C = \frac{1}{2\left(\frac{x^5+x^3+1}{x^5}\right)^2} + C = \frac{x^{10}}{2\left(x^5+x^3+1\right)^2} + C$

The value of $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$ is equal to (where $C$ is an arbitrary constant.)

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