If int {frac{{({x^2} - 1),dx}}{{({x^4} + 3{x^ | vedclass

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(B) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{\left( {{x^2} + \frac{1}{{{x^2}}} + 3} ight){{	an }^{ - 1

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

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(B) Divide the numerator and denominator by $x^2$: $\int {\frac{{1 - \frac{1}{{{x^2}}}}}{{\left( {{x^2} + \frac{1}{{{x^2}}} + 3} ight){{	an }^{ - 1}}\left( {x + \frac{1}{x}} ight)}}} dx$ Since $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2$,the integral becomes: $\int {\frac{{(1 - \frac{1}{x^2}) dx}}{{\left( (x + \frac{1}{x})^2 + 1 ight) 	an^{-1}(x + \frac{1}{x})}}}$ Let $u = 	an^{-1}(x + \frac{1}{x})$. Then $du = \frac{1}{1 + (x + \frac{1}{x})^2} \cdot (1 - \frac{1}{x^2}) dx$. The integral simplifies to $\int \frac{du}{u} = \ln|u| + C$. Substituting back,we get $\ln|	an^{-1}(x + \frac{1}{x})| + C$. Thus,$f(x) = 	an^{-1}(x + \frac{1}{x})$.

If $\int {\frac{{({x^2} - 1)\,dx}}{{({x^4} + 3{x^2} + 1)\,{{\tan }^{ - 1}}\left( {\frac{{{x^2} + 1}}{x}} \right)}}} = \ln | f(x) | + C$,then $f(x)$ is:

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