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

Explanation

To solve the integral \(\int \sqrt{\frac{a+x}{a-x}} \, dx\), we will use a trigonometric substitution. Let's go through the solution step by step.

### Step 1: Substitution
Let \( x = a \cos(2\theta) \). Then, we differentiate both sides to find \(dx\):
\[ dx = -2a \sin(2\theta) \, d\theta \]

### Step 2: Substitute in the Integral
Now, substitute \(x\) and \(dx\) into the integral:
\[ \int \sqrt{\frac{a + a \cos(2\theta)}{a - a \cos(2\theta)}} \, (-2a \sin(2\theta) \, d\theta) \]
This simplifies to:
\[ -2a \int \sqrt{\frac{a(1 + \cos(2\theta))}{a(1 - \cos(2\theta))}} \sin(2\theta) \, d\theta \]
Cancelling \(a\) gives:
\[ -2a \int \sqrt{\frac{1 + \cos(2\theta)}{1 - \cos(2\theta)}} \sin(2\theta) \, d\theta \]

### Step 3: Simplify the Square Root
Using the identity \(1 + \cos(2\theta) = 2\cos^2(\theta)\) and \(1 - \cos(2\theta) = 2\sin^2(\theta)\), we have:
\[ \sqrt{\frac{1 + \cos(2\theta)}{1 - \cos(2\theta)}} = \sqrt{\frac{2\cos^2(\theta)}{2\sin^2(\theta)}} = \frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta) \]
Thus, the integral becomes:
\[ -2a \int \cot(\theta) \sin(2\theta) \, d\theta \]

### Step 4: Simplify \(\sin(2\theta)\)
Recall that \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\):
\[ -2a \int \cot(\theta) \cdot 2\sin(\theta)\cos(\theta) \, d\theta = -4a \int \cos(\theta) \, d\theta \]

### Step 5: Integrate
The integral of \(\cos(\theta)\) is:
\[ -4a \sin(\theta) + C \]

### Step 6: Back Substitute
We need to back substitute for \(\theta\). Recall:
\[ \cos(2\theta) = \frac{x}{a} \implies 2\theta = \cos^{-1}\left(\frac{x}{a}\right) \implies \theta = \frac{1}{2} \cos^{-1}\left(\frac{x}{a}\right) \]
Using the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\[ \sin^2(\theta) = 1 - \left(\frac{x}{a}\right)^2 \implies \sin(\theta) = \sqrt{1 - \frac{x^2}{a^2}} \]
Thus, substituting back gives:
\[ -4a \sqrt{1 - \frac{x^2}{a^2}} + C \]

### Final Answer
The final result of the integral is:
\[ -4a \sqrt{1 - \frac{x^2}{a^2}} + C \]