A) \[\frac{1}{2}\ell n\left[ p-\sqrt{{{p}^{2}}-1} \right]\]
B) \[\frac{1}{2}{{\cos }^{-1}}p+\frac{1}{2}{{\sec }^{-1}}p\]
C) \[\ell n\sqrt{p+\sqrt{{{p}^{2}}-1}}-\frac{1}{2}{{\sec }^{-1}}p\]
D) None of the above.
Correct Answer: C
Solution :
[c] \[I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp}\] \[=\frac{1}{2}\int{\frac{p-1}{p\sqrt{(p+1)(p-1)}}dp}\] \[=\frac{1}{2}\int{\frac{pdp}{p\sqrt{{{p}^{2}}-1}}-\frac{1}{2}\int{\frac{dp}{p\sqrt{{{p}^{2}}-1}}}}\] \[=\frac{1}{2}\int{\frac{dp}{\sqrt{{{p}^{2}}-1}}-\frac{1}{2}\int{\frac{dp}{p\sqrt{{{p}^{2}}-1}}}}\] \[=\frac{1}{2}\log {{\,}_{e}}\left( p+\sqrt{{{p}^{2}}-1} \right)-\frac{1}{2}{{\sec }^{-1}}p\] \[\Rightarrow f(p)=\log \sqrt{p+\sqrt{{{p}^{2}}-1}}-\frac{1}{2}{{\sec }^{-1}}p\]
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