A) \[\frac{1}{2}\log \left( \frac{{{x}^{2}}+x+1}{{{x}^{2}}-x+1} \right)+c\]
B) \[\frac{1}{2}\log \left( \frac{{{x}^{2}}-x-1}{{{x}^{2}}+x+1} \right)+c\]
C) \[\log \left( \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right)+c\]
D) \[\frac{1}{2}\log \left( \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right)+c\]
Correct Answer: D
Solution :
The given function can be written as \[\int_{{}}^{{}}{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)}{{{\left( x+\frac{1}{x} \right)}^{2}}-1}}\,dx\] Put \[x+\frac{1}{x}=t\Rightarrow \left( 1-\frac{1}{{{x}^{2}}} \right)\,dx=dt,\] then it reduces to \[\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}-1}}=\frac{1}{2}\log \left| \frac{t-1}{t+1} \right|+c\] \[=\frac{1}{2}\log \left( \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} \right)+c=\frac{1}{2}\log \left( \frac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1} \right)+c.\]
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