A) \[\frac{{{x}^{2}}}{\sqrt{{{x}^{2}}-1}}+C\]
B) \[-\frac{{{x}^{2}}}{\sqrt{{{x}^{2}}-1}}+C\]
C) \[\frac{\sqrt{{{x}^{2}}-1}}{{{x}^{2}}}+C\]
D) \[-\frac{\sqrt{{{x}^{2}}-1}}{{{x}^{2}}}+C\]
Correct Answer: D
Solution :
Let \[I=\int_{{}}^{{}}{\frac{{{x}^{2}}-2}{{{x}^{3}}\sqrt{x-1}}dx}\] \[=\int_{{}}^{{}}{\frac{{{x}^{2}}}{{{x}^{3}}\sqrt{x-1}}dx}-2\int_{{}}^{{}}{\frac{1}{{{x}^{3}}\sqrt{{{x}^{2}}-1}}}dx\] \[=\int_{{}}^{{}}{\frac{1}{x\sqrt{{{x}^{2}}-1}}dx}-2\int_{{}}^{{}}{\frac{1}{{{\sec }^{3}}\theta \tan \theta }}\sec \theta \tan \theta d\theta \] \[\text{ }\!\![\!\!\text{ where,x = sec }\!\!\theta\!\!\text{ }\!\!]\!\!\text{ }\] \[={{\sec }^{-1}}x-\int_{{}}^{{}}{(1+\cos 2\theta )}d\theta \] \[={{\sec }^{-1}}x-\left( \theta +\frac{\sin 2\theta }{2} \right)+C\] \[={{\sec }^{-1}}x-{{\sec }^{-1}}x-\frac{\sqrt{{{x}^{2}}-1}}{{{x}^{2}}}+C\] \[\therefore =-\frac{\sqrt{{{x}^{2}}-1}}{{{x}^{2}}}+C\]
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