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

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Question

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

Vedclass · Accepted Answer

$\frac{1}{2}\log \left| \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} \right| + c$

Vedclass · Answer

(B) Let $I = \int {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} \,dx$.Divide the numerator and denominator by $x^2$:$I = \int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + 1 + \frac{1}{{{x^2}}}}} \,dx} = \int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 1}}} \,dx$.Let $t = x + \frac{1}{x}$,then $dt = \left( {1 - \frac{1}{{{x^2}}}} \right)dx$.Substituting these into the integral:$I = \int {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right| + c$.Substituting $t = x + \frac{1}{x}$ back:$I = \frac{1}{2}\log \left| {\frac{{x + \frac{1}{x} - 1}}{{x + \frac{1}{x} + 1}}} \right| + c = \frac{1}{2}\log \left| {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right| + c$.

$\int {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} \,dx$ is equal to

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