int frac{3 x^9+7 x^8}{left(x^2+2 x+5 x^8right | vedclass

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(B) Let $I = \int \frac{3 x^9+7 x^8}{(x^2+2 x+5 x^8)^2} dx$. Divide the numerator and denominator by $x^{16}$ inside the square: $I = \int \frac{3 x^9

Vedclass · Accepted Answer

$\frac{x^7}{2\left(5 x^7+x+2\right)}+c$

Vedclass · Answer

(B) Let $I = \int \frac{3 x^9+7 x^8}{(x^2+2 x+5 x^8)^2} dx$. Divide the numerator and denominator by $x^{16}$ inside the square: $I = \int \frac{3 x^9+7 x^8}{x^{16} (x^{-6} + 2x^{-7} + 5)^2} dx = \int \frac{3 x^{-7} + 7 x^{-8}}{(x^{-6} + 2x^{-7} + 5)^2} dx$. Let $t = x^{-6} + 2x^{-7} + 5$. Then $dt = (-6x^{-7} - 14x^{-8}) dx = -2(3x^{-7} + 7x^{-8}) dx$. So,$(3x^{-7} + 7x^{-8}) dx = -\frac{1}{2} dt$. Substituting this into the integral: $I = \int \frac{-1/2}{t^2} dt = -\frac{1}{2} \int t^{-2} dt = -\frac{1}{2} (\frac{t^{-1}}{-1}) + C = \frac{1}{2t} + C$. Substituting $t$ back: $I = \frac{1}{2(x^{-6} + 2x^{-7} + 5)} + C = \frac{1}{2(\frac{x+2+5x^7}{x^7})} + C = \frac{x^7}{2(5x^7+x+2)} + C$.

$\int \frac{3 x^9+7 x^8}{\left(x^2+2 x+5 x^8\right)^2} d x=$

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