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

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(C) 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 + \

Vedclass · Accepted Answer

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

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(C) 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}} ight)}^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}} = {{	an }^{ - 1}}t + c} $ Substituting back $t = x - \frac{1}{x}$: $= {	an ^{ - 1}}\left( {x - \frac{1}{x}} ight) + c = {	an ^{ - 1}}\left( {\frac{{{x^2} - 1}}{x}} ight) + c$.

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

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