The primitive of frac{3x^4 - 1}{(x^4 + x + 1) | vedclass

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Question

(B) Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Divide the numerator and denominator by $x^4$: $I = \int \frac{3 - x^{-4}}{(1 + x^{-3} + x^{-4

Vedclass · Accepted Answer

$-\frac{x}{x^4 + x + 1} + c$

Vedclass · Answer

(B) Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Divide the numerator and denominator by $x^4$: $I = \int \frac{3 - x^{-4}}{(1 + x^{-3} + x^{-4})^2} dx$. This does not simplify easily. Let us rewrite the integrand by dividing numerator and denominator by $x^2$: $I = \int \frac{3x^2 - x^{-2}}{(x^2 + 1 + x^{-1})^2} dx$. Let $u = x^2 + 1 + x^{-1}$. Then $du = (2x - x^{-2}) dx$. This is not matching. Let us try $u = \frac{x}{x^4 + x + 1}$. Using the quotient rule,$du = \frac{(x^4 + x + 1)(1) - x(4x^3 + 1)}{(x^4 + x + 1)^2} dx = \frac{x^4 + x + 1 - 4x^4 - x}{(x^4 + x + 1)^2} dx = \frac{1 - 3x^4}{(x^4 + x + 1)^2} dx$. Thus,$du = - \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Therefore,$I = - \int du = -u + c = -\frac{x}{x^4 + x + 1} + c$.

The primitive of $\frac{3x^4 - 1}{(x^4 + x + 1)^2}$ with respect to $x$ is:

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