A) \[\log \frac{{{x}^{2}}-1}{x}+c\]
B) \[-\log \frac{{{x}^{2}}-1}{x}+c\]
C) \[\log \frac{x}{{{x}^{2}}+1}+c\]
D) \[-\log \frac{x}{{{x}^{2}}+1}+c\]
Correct Answer: A
Solution :
\[I=\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{x({{x}^{2}}-1)}\,dx}=\int_{{}}^{{}}{\frac{1+\left( \frac{1}{{{x}^{2}}} \right)}{x-\left( \frac{1}{x} \right)}\,dx}\] Put \[x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{{{x}^{2}}})\,dx=dt\] \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{dt}{t}}=\log t+c=\log \frac{{{x}^{2}}-1}{x}+c\].You need to login to perform this action.
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