Evaluate : `int sqrt((a+x)/(a-x)) dx`

Answer

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

### Step 1: Rationalize the Integrand
We start by rationalizing the integrand:
\[ I = \int \sqrt{\frac{a+x}{a-x}} \, dx = \int \frac{\sqrt{(a+x)(a+x)}}{\sqrt{(a-x)(a+x)}} \, dx = \int \frac{a+x}{\sqrt{a^2 - x^2}} \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ I = \int \frac{a}{\sqrt{a^2 - x^2}} \, dx + \int \frac{x}{\sqrt{a^2 - x^2}} \, dx \]
Let’s denote these two integrals as \( I_1 \) and \( I_2 \):
\[ I_1 = \int \frac{a}{\sqrt{a^2 - x^2}} \, dx, \quad I_2 = \int \frac{x}{\sqrt{a^2 - x^2}} \, dx \]

### Step 3: Evaluate \( I_1 \)
The integral \( I_1 \) can be evaluated using the standard integral formula:
\[ I_1 = a \sin^{-1}\left(\frac{x}{a}\right) + C_1 \]

### Step 4: Evaluate \( I_2 \)
For \( I_2 \), we use the substitution \( t = a^2 - x^2 \):
\[ dt = -2x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{2x} \]
Thus, we have:
\[ I_2 = \int \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right) = -\frac{1}{2} \int t^{-\frac{1}{2}} \, dt \]
This evaluates to:
\[ I_2 = -\frac{1}{2} \cdot 2\sqrt{t} + C_2 = -\sqrt{a^2 - x^2} + C_2 \]

### Step 5: Combine Results
Now, we combine the results from \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = a \sin^{-1}\left(\frac{x}{a}\right) - \sqrt{a^2 - x^2} + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = a \sin^{-1}\left(\frac{x}{a}\right) - \sqrt{a^2 - x^2} + C \]