A) \[\log ({{x}^{2}}+{{x}^{2}}+1)+c\]
B) \[\log \left| \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right|+c\]
C) \[\frac{1}{2}\log \left| \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right|+c\]
D) \[\frac{1}{2}\log \left| \frac{{{x}^{2}}+x+1}{{{x}^{2}}-x+1} \right|+c\]
Correct Answer: C
Solution :
\[\int{\frac{({{x}^{2}}-1)dx}{{{x}^{4}}+{{x}^{2}}+1}}=\int{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)dx}{\left( {{x}^{2}}+1+\frac{1}{{{x}^{2}}} \right)}}\] \[=\int{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x+\frac{1}{x} \right)}^{2}}-1}}\] Put \[x+\frac{1}{x}=t\] \[\Rightarrow \] \[\left( 1-\frac{1}{{{x}^{2}}} \right)dx=dt\] \[\therefore \] \[\int{\frac{dt}{{{t}^{2}}-1}=\frac{1}{2}\log \left| \frac{t-1}{t+1} \right|+c}\] \[=\frac{1}{2}\log \left| \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} \right|+c\] \[=\frac{1}{2}\log \left| \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right|+c\]
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