A) \[\frac{1}{2}{{\sec }^{-1}}{{x}^{2}}+k\]
B) \[\log x\sqrt{{{x}^{4}}-1}+k\]
C) \[x\log \sqrt{{{x}^{4}}-1}+k\]
D) \[\log \sqrt{{{x}^{4}}-1}+k\]
Correct Answer: A
Solution :
\[I=\int_{{}}^{{}}{\frac{dx}{x\sqrt{{{x}^{4}}-1}}}\] Put \[{{x}^{2}}=t\Rightarrow 2x\,dx=dt\Rightarrow dx=\frac{dt}{2x}=\frac{dt}{2\sqrt{t}}\] \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{dt}{2t\sqrt{{{t}^{2}}-1}}}=\frac{1}{2}{{\sec }^{-1}}t+k=\frac{1}{2}{{\sec }^{-1}}{{x}^{2}}+k\]
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