Evaluate: `int(ddx)/(sqrt(1+x^2)+sqrt((1+x^2)^3))`

Explanation

To evaluate the integral

\[ I = \int \frac{x \, dx}{\sqrt{1+x^2} + \sqrt{(1+x^2)^3}} \]

we can follow these steps:

### Step 1: Simplify the Integral
First, we can factor out \(\sqrt{1+x^2}\) from the denominator:

\[ I = \int \frac{x \, dx}{\sqrt{1+x^2} \left(1 + \sqrt{(1+x^2)^2}\right)} \]

Notice that \(\sqrt{(1+x^2)^2} = 1+x^2\). Thus, we can rewrite the integral as:

\[ I = \int \frac{x \, dx}{\sqrt{1+x^2} \left(1 + (1+x^2)\right)} = \int \frac{x \, dx}{\sqrt{1+x^2} (2+x^2)} \]

### Step 2: Use Substitution
Let us use the substitution \(t = \sqrt{1+x^2}\). Then, we have:

\[ t^2 = 1 + x^2 \implies x^2 = t^2 - 1 \implies x = \sqrt{t^2 - 1} \]

Differentiating both sides gives:

\[ dx = \frac{x}{\sqrt{1+x^2}} \, dt = \frac{\sqrt{t^2 - 1}}{t} \, dt \]

### Step 3: Substitute into the Integral
Now substitute \(x\) and \(dx\) into the integral:

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

This simplifies to:

\[ I = \int \frac{(t^2 - 1)}{t^2 (1 + t^2)} \, dt \]

### Step 4: Simplify Further
Now, we can break this integral into two parts:

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

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

1. \(\int \frac{1}{1+t^2} \, dt = \tan^{-1}(t)\)
2. \(\int \frac{1}{t^2} \, dt = -\frac{1}{t}\)

Thus, we have:

\[ I = \tan^{-1}(t) - \frac{1}{t} + C \]

### Step 6: Substitute Back
Now substitute back \(t = \sqrt{1+x^2}\):

\[ I = \tan^{-1}(\sqrt{1+x^2}) - \frac{1}{\sqrt{1+x^2}} + C \]

### Final Answer
The final result of the integral is:

\[ I = \tan^{-1}(\sqrt{1+x^2}) - \frac{1}{\sqrt{1+x^2}} + C \]