A) \[\frac{1}{\sqrt{7}}{{\tan }^{-1}}\left( \frac{4x+1}{\sqrt{7}} \right)+c\]
B) \[\frac{1}{2\sqrt{7}}{{\tan }^{-1}}\left( \frac{4x+1}{\sqrt{7}} \right)+c\]
C) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{4x+1}{\sqrt{7}} \right)+c\]
D) None of these
Correct Answer: D
Solution :
\[I=\int_{{}}^{{}}{\frac{dx}{2{{x}^{2}}+x+1}}=\int_{{}}^{{}}{\frac{dx}{2\left( {{x}^{2}}+\frac{x}{2}+\frac{1}{2} \right)}}\] \[=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+\frac{x}{2}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}}}=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{{{\left( x+\frac{1}{4} \right)}^{2}}+{{\left( \frac{\sqrt{7}}{4} \right)}^{2}}}}\] \[=\frac{1}{2}\frac{1}{\frac{\sqrt{7}}{4}}{{\tan }^{-1}}\frac{[x+(1/4)]}{\sqrt{7}/4}\]\[=\frac{2}{\sqrt{7}}{{\tan }^{-1}}\frac{(4x+1)}{\sqrt{7}}+C\].
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