The integral int frac{3x^{13} + 2x^{11}}{(2x^ | vedclass

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(B) We have the integral $I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} dx$. Divide the numerator and denominator by $x^{16}$ inside the inte

Vedclass · Accepted Answer

$\frac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$

Vedclass · Answer

(B) We have the integral $I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} dx$. Divide the numerator and denominator by $x^{16}$ inside the integral: $I = \int \frac{\frac{3x^{13} + 2x^{11}}{x^{16}}}{(\frac{2x^4 + 3x^2 + 1}{x^4})^4} dx = \int \frac{3x^{-3} + 2x^{-5}}{(2 + 3x^{-2} + x^{-4})^4} dx$. Let $t = 2 + 3x^{-2} + x^{-4}$. Then $dt = (-6x^{-3} - 4x^{-5}) dx = -2(3x^{-3} + 2x^{-5}) dx$. So,$(3x^{-3} + 2x^{-5}) dx = -\frac{1}{2} dt$. Substituting these into the integral: $I = \int \frac{-\frac{1}{2} dt}{t^4} = -\frac{1}{2} \int t^{-4} dt = -\frac{1}{2} \left( \frac{t^{-3}}{-3} \right) + C = \frac{1}{6t^3} + C$. Substituting $t$ back: $I = \frac{1}{6(2 + 3x^{-2} + x^{-4})^3} + C = \frac{1}{6(\frac{2x^4 + 3x^2 + 1}{x^4})^3} + C = \frac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$.

The integral $\int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} dx$ is equal to (where $C$ is a constant of integration)

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