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$If \int \frac{(x - 1)^{2}}{x^{4} + x^{2} + 1} d x = f (x) + C, then the value of lim_{x \to \infty} f (x) is$

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Correct option is A

∫(x−1)2x4+x2+1dx =∫x2+1x4+x2+1dx−∫2xx4+x2+1dx =∫1+1x2x−1x2+3dx−∫2xx2+122+34dx =13tan−1⁡x−1x3−23tan−1⁡2x2+13+C =f(x)+C∴ limx→∞ f(x)=13⋅π2−23⋅π2=−π23

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If ∫(x−1)2x4+x2+1dx=f(x)+C, then the value of limx→∞ f(x) is

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$If \int \frac{(x - 1)^{2}}{x^{4} + x^{2} + 1} d x = f (x) + C, then the value of lim_{x \to \infty} f (x) is$