`int[(x)/(sqrt(x^(2)+a^(2))+sqrt(x^(2)-a^(2)))]dx`

Answer

To solve the integral

\[ I = \int \frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} \, dx, \]

we can follow these steps:

### Step 1: Substitution
Let \( x^2 = t \). Then, \( dx = \frac{1}{2\sqrt{t}} \, dt \).

### Step 2: Rewrite the Integral
Substituting \( x^2 = t \) into the integral gives:

\[ I = \int \frac{\sqrt{t}}{\sqrt{t + a^2} + \sqrt{t - a^2}} \cdot \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \int \frac{1}{\sqrt{t + a^2} + \sqrt{t - a^2}} \, dt. \]

### Step 3: Rationalization
To simplify the integral, we rationalize the denominator:

\[ \frac{1}{\sqrt{t + a^2} + \sqrt{t - a^2}} \cdot \frac{\sqrt{t + a^2} - \sqrt{t - a^2}}{\sqrt{t + a^2} - \sqrt{t - a^2}} = \frac{\sqrt{t + a^2} - \sqrt{t - a^2}}{(t + a^2) - (t - a^2)} = \frac{\sqrt{t + a^2} - \sqrt{t - a^2}}{2a^2}. \]

### Step 4: Substitute Back
Now, substituting back into the integral, we have:

\[ I = \frac{1}{2} \int \frac{\sqrt{t + a^2} - \sqrt{t - a^2}}{2a^2} \, dt = \frac{1}{4a^2} \int \left( \sqrt{t + a^2} - \sqrt{t - a^2} \right) dt. \]

### Step 5: Integrate
Now we can integrate each term separately:

1. For \( \int \sqrt{t + a^2} \, dt \):
\[ \int \sqrt{t + a^2} \, dt = \frac{2}{3} (t + a^2)^{3/2} + C_1. \]

2. For \( \int \sqrt{t - a^2} \, dt \):
\[ \int \sqrt{t - a^2} \, dt = \frac{2}{3} (t - a^2)^{3/2} + C_2. \]

### Step 6: Combine Results
Combining these results gives:

\[ I = \frac{1}{4a^2} \left( \frac{2}{3} (t + a^2)^{3/2} - \frac{2}{3} (t - a^2)^{3/2} \right) + C. \]

### Step 7: Substitute Back for \( t \)
Replacing \( t \) back with \( x^2 \):

\[ I = \frac{1}{6a^2} \left( (x^2 + a^2)^{3/2} - (x^2 - a^2)^{3/2} \right) + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ I = \frac{1}{6a^2} \left( (x^2 + a^2)^{3/2} - (x^2 - a^2)^{3/2} \right) + C. \]