`int(sqrt((a+x)/(a-x))+sqrt((a-x)/(a+x)))dx " is equal to "`

Options

A. `2sin^(-1)(x//a)+c`
B. `2asin^(-1)(x//a)+c`
C. `2cos^(-1)(x//a)+c`
D. `2acos^(-1)(x//a)+c`

Explanation

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

### Step 1: Rewrite the integral
We start with the given integral:
\[ I = \int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx \]

### Step 2: Combine the terms
We can combine the two square root terms:
\[ I = \int \left( \sqrt{\frac{(a+x)^2 + (a-x)^2}{(a-x)(a+x)}} \right) dx \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:
\[ (a+x)^2 + (a-x)^2 = (a^2 + 2ax + x^2) + (a^2 - 2ax + x^2) = 2a^2 + 2x^2 \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sqrt{2(a^2 + x^2)}}{\sqrt{a^2 - x^2}} dx \]

### Step 4: Factor out constants
We can factor out the constant \( \sqrt{2} \):
\[ I = \sqrt{2} \int \frac{\sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2}} dx \]

### Step 5: Use trigonometric substitution
To solve the integral, we can use the substitution \( x = a \sin \theta \). Then, \( dx = a \cos \theta d\theta \) and the limits change accordingly:
\[ I = \sqrt{2} \int \frac{\sqrt{a^2 + a^2 \sin^2 \theta}}{\sqrt{a^2 - a^2 \sin^2 \theta}} a \cos \theta d\theta \]

### Step 6: Simplify the integral
This simplifies to:
\[ I = \sqrt{2} \int \frac{a \sqrt{a^2 (1 + \sin^2 \theta)}}{a \sqrt{a^2 (1 - \sin^2 \theta)}} a \cos \theta d\theta \]
\[ = \sqrt{2} \int \frac{\sqrt{1 + \sin^2 \theta}}{\sqrt{1 - \sin^2 \theta}} a \cos \theta d\theta \]
\[ = \sqrt{2} \int \frac{\sqrt{1 + \sin^2 \theta}}{\cos \theta} a \cos \theta d\theta \]
\[ = \sqrt{2} a \int \sqrt{1 + \sin^2 \theta} d\theta \]

### Step 7: Solve the integral
The integral \( \int \sqrt{1 + \sin^2 \theta} d\theta \) can be solved using standard techniques or looked up in integral tables.

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