`int sqrt((9+x)/(9-x))dx`

Answer

To solve the integral \( \int \sqrt{\frac{9+x}{9-x}} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \sqrt{\frac{9+x}{9-x}} \, dx \]

### Step 2: Simplify the Expression
Multiply the numerator and denominator by \( \sqrt{9+x} \):
\[ I = \int \frac{(9+x)}{\sqrt{(9-x)(9+x)}} \, dx = \int \frac{(9+x)}{\sqrt{81 - x^2}} \, dx \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ I = \int \frac{9}{\sqrt{81 - x^2}} \, dx + \int \frac{x}{\sqrt{81 - x^2}} \, dx \]
Let’s denote these two integrals as \( I_1 \) and \( I_2 \):
\[ I_1 = \int \frac{9}{\sqrt{81 - x^2}} \, dx \]
\[ I_2 = \int \frac{x}{\sqrt{81 - x^2}} \, dx \]

### Step 4: Solve \( I_1 \)
For \( I_1 \):
\[ I_1 = 9 \int \frac{1}{\sqrt{81 - x^2}} \, dx \]
The integral \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C \) where \( a = 9 \):
\[ I_1 = 9 \sin^{-1}\left(\frac{x}{9}\right) + C \]

### Step 5: Solve \( I_2 \)
For \( I_2 \), we use the substitution \( t = 81 - x^2 \):
\[ dt = -2x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{2x} \]
Substituting in \( I_2 \):
\[ I_2 = \int \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right) = -\frac{1}{2} \int t^{-1/2} \, dt \]
The integral \( \int t^{-1/2} \, dt = 2\sqrt{t} + C \):
\[ I_2 = -\frac{1}{2} \cdot 2\sqrt{t} + C = -\sqrt{81 - x^2} + C \]

### Step 6: Combine \( I_1 \) and \( I_2 \)
Now we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = 9 \sin^{-1}\left(\frac{x}{9}\right) - \sqrt{81 - x^2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sqrt{\frac{9+x}{9-x}} \, dx = 9 \sin^{-1}\left(\frac{x}{9}\right) - \sqrt{81 - x^2} + C \]