int frac{2 x^2-1}{x^4-x^2-20} d x | vedclass

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Question

(C) Let $I = \int \frac{2 x^2-1}{x^4-x^2-20} d x$. Substitute $x^2 = t$,then the integrand becomes $\frac{2 t-1}{t^2-t-20} = \frac{2 t-1}{(t-5)(t+4)}$

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{5}} \log \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}ight|+\frac{1}{2} 	an ^{-1}\left(\frac{x}{2}ight)+c$

Vedclass · Answer

(C) Let $I = \int \frac{2 x^2-1}{x^4-x^2-20} d x$. Substitute $x^2 = t$,then the integrand becomes $\frac{2 t-1}{t^2-t-20} = \frac{2 t-1}{(t-5)(t+4)}$. Using partial fractions,let $\frac{2 t-1}{(t-5)(t+4)} = \frac{A}{t-5} + \frac{B}{t+4}$. Then $2t-1 = A(t+4) + B(t-5)$. For $t=5$,$9 = 9A \implies A=1$. For $t=-4$,$-9 = -9B \implies B=1$. Thus,$I = \int \left( \frac{1}{x^2-5} + \frac{1}{x^2+4} ight) d x$. Using the standard integrals $\int \frac{1}{x^2-a^2} d x = \frac{1}{2a} \log \left| \frac{x-a}{x+a} ight| + c$ and $\int \frac{1}{x^2+a^2} d x = \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) + c$,we get: $I = \frac{1}{2\sqrt{5}} \log \left| \frac{x-\sqrt{5}}{x+\sqrt{5}} ight| + \frac{1}{2} 	an^{-1} \left( \frac{x}{2} ight) + c$.

$ \int \frac{2 x^2-1}{x^4-x^2-20} d x $

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