A) \[\sqrt{\frac{x+1}{x-1}}+c\]
B) \[\sqrt{\frac{x-1}{x+1}}+c\]
C) \[\sqrt{\frac{1}{x+1}}+c\]
D) None of these
Correct Answer: B
Solution :
Let\[I=\int{\frac{1}{(x+1)\sqrt{{{x}^{2}}-1}}dx}\] Put \[x+1=\frac{1}{t}\] \[\Rightarrow \] \[dx=-\frac{1}{{{t}^{2}}}dt\], then \[I=\int{\frac{1}{\frac{1}{t}{{\left( \frac{1}{t}-1 \right)}^{2}}}\left( \frac{-1}{{{t}^{2}}} \right)dt}\] \[\Rightarrow \] \[I=-\int{\frac{dt}{\sqrt{1-dt}}=-\int{{{(1-2t)}^{-1/2}}dt}}\] \[=-\frac{{{(1-2t)}^{1/2}}}{(-2)\left( \frac{1}{2} \right)}+c\] \[\Rightarrow \] \[I=\sqrt{1-2t}+c\] \[\Rightarrow \] \[I=\sqrt{1-\frac{2}{x+1}}+c\] \[\Rightarrow \] \[I=\sqrt{\frac{x-1}{x+1}}+c\]
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