Question

\(\int\frac{dx}{(x+1)\sqrt{x^2-1}}\) is equal to:
[\(c\) is an arbitrary constant]

• A. \(\sqrt{\frac{x+1}{x-1}}+c\)
• B. \(\sqrt{\frac{x}{x-1}}+c\)
• C. \(\sqrt{\frac{x-1}{x+1}}+c\)
• D. \(\sqrt{\frac{x}{x+1}}+c\)

Hint:

Write \(\sqrt{x^2-1}=\sqrt{(x-1)(x+1)}\) and substitute \(t=\frac{x-1}{x+1}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We rewrite \(\sqrt{x^2-1}\) as \(\sqrt{(x-1)(x+1)}\) and then use a substitution that simplifies the ratio \(\frac{x-1}{x+1}\).

Step 2: Rewrite:
\[ \frac{1}{(x+1)\sqrt{(x-1)(x+1)}}=\frac{1}{(x+1)^{2}}\cdot\sqrt{\frac{x+1}{x-1}} \] because \((x+1)\sqrt{x+1}\sqrt{x-1}=(x+1)^2\sqrt{\frac{x-1}{x+1}}\).

Step 3: Substitute:
Let \(t=\frac{x-1}{x+1}=1-\frac{2}{x+1}\). Then \(dt=\frac{2}{(x+1)^2}dx\), so \(\frac{dx}{(x+1)^2}=\frac{dt}{2}\).

Step 4: Integrate:
\[ I=\int\frac{1}{\sqrt t}\cdot\frac{dt}{2}=\frac12\cdot 2\sqrt t+c=\sqrt t+c \]

Step 5: Back-substitute:
\[ I=\sqrt{\frac{x-1}{x+1}}+c \]

Step 6: Check by differentiating:
\(\frac{d}{dx}\sqrt{\frac{x-1}{x+1}}=\frac{1}{2}\sqrt{\frac{x+1}{x-1}}\cdot\frac{2}{(x+1)^2}=\frac{1}{(x+1)^{3/2}(x-1)^{1/2}}\), which equals the integrand. Option 1 is the reciprocal of the correct ratio, so its derivative has the wrong sign. Options 2 and 4 do not give the integrand either.

Final Answer:
The integral equals \(\sqrt{\frac{x-1}{x+1}}+c\), option 3. \[ \boxed{\sqrt{\frac{x-1}{x+1}}+c} \]