A) \[\log \left| x+\frac{1}{x}+\sqrt{{{x}^{2}}+\frac{1}{{{x}^{2}}}+3} \right|+C\]
B) \[\log \left| x-\frac{1}{x}+\sqrt{{{x}^{2}}+\frac{1}{{{x}^{2}}}-3} \right|+C\]
C) \[\log \left| x+\sqrt{{{x}^{2}}+3} \right|+C\]
D) None of these
Correct Answer: A
Solution :
\[\int{\frac{({{x}^{2}}-1)}{x\sqrt{{{x}^{4}}+3{{x}^{2}}+1}}}\,dx=\int{\frac{({{x}^{2}}-1)}{{{x}^{2}}\sqrt{{{x}^{2}}+3+\frac{1}{{{x}^{2}}}}}\,dx}\] \[=\,\,\,\int{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)}{{{\left( x+\frac{1}{x} \right)}^{2}}+1}}\,dx=\int{\frac{dz}{\sqrt{{{z}^{2}}+1}}}\] \[\left[ Putting\,\,x+\frac{1}{x}=z\,\,\Rightarrow \,\,\left( 1-\frac{1}{{{x}^{2}}} \right)dx=dz \right]\] \[=\,\,\,\log \left| z+\sqrt{{{z}^{2}}+1} \right|+C\] \[=\,\,\,\log \left| x+\frac{1}{x}+\sqrt{{{x}^{2}}+\frac{1}{{{x}^{2}}}+3} \right|+C\]
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