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

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(D) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + 1 + \frac{1}{{{x^2}}}}}\,dx} = \int {\frac{{1 - \fra

Vedclass · Accepted Answer

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

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

$\int {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}\,dx = }$

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