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

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Question

(D) Let $I = \int \frac{3 x^{13}+2 x^{11}}{\left(2 x^4+3 x^2+1\right)^4} d x$. Divide the numerator and denominator by $x^{16}$ inside the integral: $

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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