Q135 / 200 · Manipal (MET) 2010 · Mathematics
Evaluate $\int \frac{1}{(x+1) \sqrt{x^2-1}} d x$
- A$\sqrt{\frac{x+1}{x-1}}+c$
- Correct answerB$\sqrt{\frac{x-1}{x+1}}+c$
- C$\sqrt{\frac{1}{x+1}}+c$
- DNone of these
Solution
Let $I=\int \frac{1}{(x+1) \sqrt{x^2-1}} d x$
Put, $x+1=\frac{1}{t}$
$\Rightarrow d x=-\frac{1}{t^2} d t$, then
$I=\int \frac{1}{\frac{1}{t} \sqrt{\left(\frac{1}{t}-1\right)^2-1}}\left(\frac{-1}{t^2}\right) d t$
$\Rightarrow I=-\int \frac{d t}{\sqrt{1-2 t}}=-\int(1-2 t)^{-1 / 2} d t$
$=-\frac{(1-2 t)^{1 / 2}}{(-2)\left(\frac{1}{2}\right)}+c$
$\Rightarrow I=\sqrt{1-2 t}+c$
$\Rightarrow I=\sqrt{1-\frac{2}{x+1}}+c$
$\Rightarrow I=\sqrt{\frac{x-1}{x+1}}+c$
Hence, the correct answer is option: 'B'.