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