`int(1)/((x+1)sqrt(x-1))dx=`

Options

A. `(1)/(2sqrt2)tan^(-1)((x-1)/(2))+c`
B. `sqrt2 tan^(-1)sqrt((x-1)/(2))+c`
C. `(1)/(2sqrt2)log(x+sqrt(x-1))+c`
D. `-sqrt2log(x+1+sqrt(x-1))+c`

Answer

To solve the integral \( \int \frac{1}{(x+1)\sqrt{x-1}} \, dx \), we will use a substitution method.

### Step-by-Step Solution:

1. **Substitution**: Let \( y^2 = x - 1 \). Then, we differentiate both sides to find \( dx \):
\[ dx = 2y \, dy \]
From the substitution, we can express \( x \) in terms of \( y \):
\[ x = y^2 + 1 \]

2. **Rewrite the Integral**: Substitute \( x \) and \( dx \) into the integral:
\[ \int \frac{1}{(y^2 + 1 + 1)\sqrt{y^2}} \cdot 2y \, dy = \int \frac{2y}{(y^2 + 2)y} \, dy \]
This simplifies to:
\[ \int \frac{2}{y^2 + 2} \, dy \]

3. **Factor Out Constants**: We can factor out the constant:
\[ 2 \int \frac{1}{y^2 + 2} \, dy \]

4. **Recognize the Standard Integral**: The integral \( \int \frac{1}{y^2 + a^2} \, dy \) is known to be:
\[ \frac{1}{a} \tan^{-1}\left(\frac{y}{a}\right) + C \]
Here, \( a^2 = 2 \) implies \( a = \sqrt{2} \).

5. **Integrate**: Thus, we have:
\[ 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{y}{\sqrt{2}}\right) + C = \frac{2}{\sqrt{2}} \tan^{-1}\left(\frac{y}{\sqrt{2}}\right) + C \]
Simplifying \( \frac{2}{\sqrt{2}} \) gives us \( \sqrt{2} \):
\[ \sqrt{2} \tan^{-1}\left(\frac{y}{\sqrt{2}}\right) + C \]

6. **Back Substitute**: Recall that \( y = \sqrt{x - 1} \):
\[ \sqrt{2} \tan^{-1}\left(\frac{\sqrt{x - 1}}{\sqrt{2}}\right) + C \]

### Final Answer:
\[ \int \frac{1}{(x+1)\sqrt{x-1}} \, dx = \sqrt{2} \tan^{-1}\left(\frac{\sqrt{x - 1}}{\sqrt{2}}\right) + C \]