A) \[\frac{1}{6}(3\pi -4)\]
B) \[\frac{1}{6}(3-4\pi )\]
C) \[\frac{1}{6}(3\pi +4)\]
D) \[\frac{1}{6}(3+4\pi )\]
Correct Answer: A
Solution :
\[I=\int_{0}^{1}{\frac{{{x}^{4}}+1}{{{x}^{2}}+1}dx=\int_{0}^{1}{\frac{{{x}^{4}}-1}{{{x}^{2}}+1}dx+2\int_{0}^{1}{\frac{dx}{1+{{x}^{2}}}}}}\] Þ \[I=\int_{0}^{1}{({{x}^{2}}-1)}dx+2\int_{0}^{1}{\frac{dx}{1+{{x}^{2}}}}\] Þ \[I=\left[ \frac{{{x}^{3}}}{3}-x \right]_{0}^{1}+2\,[{{\tan }^{-1}}x]_{0}^{1}\] \[=-\frac{2}{3}+\frac{\pi }{2}=\frac{(3\pi -4)}{6}\].
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