\( \int\displaystyle\frac{dx}{(x+1)\sqrt{x^2-1}} \) is equal to:
[\( c \) is an arbitrary constant]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \sqrt{\displaystyle\frac{x-1}{x+1}}+c \)
For x > 1, \( \sqrt{x^2-1}=\sqrt{x-1}\,\sqrt{x+1} \), so that
\( \displaystyle \frac{1}{(x+1)\sqrt{x^2-1}}=\frac{1}{(x+1)^{3/2}(x-1)^{1/2}} \)
\( \displaystyle =\frac{1}{(x+1)^2}\sqrt{\frac{x+1}{x-1}} \)
Let
\( \displaystyle t=\frac{x-1}{x+1} \)
\( \displaystyle dt=\frac{(x+1)-(x-1)}{(x+1)^2}\,dx \)
\( \displaystyle =\frac{2}{(x+1)^2}\,dx \)
Then
\( \displaystyle \int\frac{dx}{(x+1)\sqrt{x^2-1}}=\int\frac{1}{\sqrt{t}}\cdot\frac{dt}{2} \)
\( \displaystyle =\frac{1}{2}\cdot 2\sqrt{t}+c \)
\( \displaystyle =\sqrt{\frac{x-1}{x+1}}+c \)
Shortcut: Check by differentiating Option (3):
\( \displaystyle \frac{d}{dx}\sqrt{\frac{x-1}{x+1}}=\frac{1}{2}\sqrt{\frac{x+1}{x-1}}\cdot\frac{2}{(x+1)^2} \)
\( \displaystyle =\frac{1}{(x+1)^{3/2}(x-1)^{1/2}} \)
\( \displaystyle =\frac{1}{(x+1)\sqrt{x^2-1}} \)