A) \[\frac{1}{3}{{\sec }^{-1}}\left( \frac{{{x}^{3}}}{4} \right)+c\]
B) \[{{\cos }^{-1}}\left( \frac{{{x}^{3}}}{4} \right)+c\]
C) \[\frac{1}{12}{{\sec }^{-1}}\left( \frac{{{x}^{3}}}{4} \right)+c\]
D) \[{{\sec }^{-1}}\left( \frac{{{x}^{3}}}{4} \right)+c\]
Correct Answer: C
Solution :
Let \[I=\int{\frac{dx}{x\sqrt{{{x}^{6}}-16}}}\] \[=\frac{1}{3}\int{\frac{3{{x}^{2}}}{{{x}^{3}}\sqrt{{{({{x}^{3}})}^{2}}-{{4}^{2}}}}}dx\] Put \[{{x}^{3}}=t\] \[\Rightarrow \] \[3{{x}^{2}}dx=dt\] \[\therefore \] \[I=\frac{1}{3}\int{\frac{dt}{t\sqrt{{{t}^{2}}-{{4}^{2}}}}}\] \[=\frac{1}{3\times 4}{{\sec }^{-1}}\left( \frac{t}{4} \right)+c\] \[=\frac{1}{12}{{\sec }^{-1}}\left( \frac{{{x}^{3}}}{4} \right)+c\]
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