`intsqrt((x+1)/(x-1))dx`

Answer

To solve the integral \( I = \int \sqrt{\frac{x+1}{x-1}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral:**
We start with the integral:
\[ I = \int \sqrt{\frac{x+1}{x-1}} \, dx \]
We can multiply and divide the integrand by \( \sqrt{x+1} \):
\[ I = \int \frac{\sqrt{x+1} \cdot \sqrt{x+1}}{\sqrt{x-1}} \, dx = \int \frac{x+1}{\sqrt{x-1}} \, dx \]

2. **Separate the Integral:**
We can separate the integral into two parts:
\[ I = \int \frac{x}{\sqrt{x-1}} \, dx + \int \frac{1}{\sqrt{x-1}} \, dx \]

3. **Substitution for the First Integral:**
For the first integral, we use the substitution \( t = x - 1 \) which implies \( x = t + 1 \) and \( dx = dt \):
\[ I_1 = \int \frac{t + 1}{\sqrt{t}} \, dt = \int \left( \sqrt{t} + \frac{1}{\sqrt{t}} \right) dt \]

4. **Integrate Each Term:**
Now we integrate each term:
\[ I_1 = \int \sqrt{t} \, dt + \int \frac{1}{\sqrt{t}} \, dt \]
The integrals can be computed as follows:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C_1 \]
\[ \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} + C_2 \]
Thus,
\[ I_1 = \frac{2}{3} t^{3/2} + 2\sqrt{t} + C \]

5. **Substituting Back:**
Replacing \( t \) back with \( x - 1 \):
\[ I_1 = \frac{2}{3} (x-1)^{3/2} + 2\sqrt{x-1} + C \]

6. **Second Integral:**
Now we compute the second integral:
\[ I_2 = \int \frac{1}{\sqrt{x-1}} \, dx \]
Using the substitution \( t = x - 1 \):
\[ I_2 = \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} + C_3 \]
Substituting back:
\[ I_2 = 2\sqrt{x-1} + C \]

7. **Combine Both Integrals:**
Finally, we combine both parts:
\[ I = I_1 + I_2 = \left( \frac{2}{3} (x-1)^{3/2} + 2\sqrt{x-1} \right) + \left( 2\sqrt{x-1} \right) \]
Simplifying:
\[ I = \frac{2}{3} (x-1)^{3/2} + 4\sqrt{x-1} + C \]

### Final Result:
Thus, the integral evaluates to:
\[ I = \frac{2}{3} (x-1)^{3/2} + 4\sqrt{x-1} + C \]