Evaluate `int (1)/((x+1)sqrt(x-2))dx`.

Answer

To evaluate the integral \(\int \frac{1}{(x+1)\sqrt{x-2}} \, dx\), we will use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let us make the substitution:
\[ x - 2 = t^2 \]
Then, differentiating both sides, we get:
\[ dx = 2t \, dt \]
Also, we can express \(x\) in terms of \(t\):
\[ x = t^2 + 2 \]

### Step 2: Rewrite the Integral
Substituting \(x\) and \(dx\) into the integral, we have:
\[ \int \frac{1}{(t^2 + 2 + 1)\sqrt{t^2}} \cdot 2t \, dt \]
This simplifies to:
\[ \int \frac{2t}{(t^2 + 3)t} \, dt \]
The \(t\) in the numerator and denominator cancels out:
\[ \int \frac{2}{t^2 + 3} \, dt \]

### Step 3: Factor Out the Constant
We can factor out the constant 2:
\[ 2 \int \frac{1}{t^2 + 3} \, dt \]

### Step 4: Use the Standard Integral Formula
The integral \(\int \frac{1}{x^2 + a^2} \, dx\) is given by:
\[ \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \(a^2 = 3\) so \(a = \sqrt{3}\). Thus, we have:
\[ 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + C \]

### Step 5: Substitute Back
Now we substitute back \(t = \sqrt{x - 2}\):
\[ \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{x - 2}}{\sqrt{3}} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{(x+1)\sqrt{x-2}} \, dx = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{x - 2}}{\sqrt{3}} \right) + C \]

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