Evaluate: `int1/(x^2sqrt(1+x^2))dx`

Explanation

To evaluate the integral

\[ \int \frac{1}{x^2 \sqrt{1+x^2}} \, dx, \]

we will use the substitution \( x = \tan \theta \).

### Step 1: Substitution
Let \( x = \tan \theta \). Then, the differential \( dx \) is given by:

\[ dx = \sec^2 \theta \, d\theta. \]

### Step 2: Substitute in the integral
Now substitute \( x \) and \( dx \) into the integral:

\[ \int \frac{1}{\tan^2 \theta \sqrt{1+\tan^2 \theta}} \sec^2 \theta \, d\theta. \]

### Step 3: Simplify the expression
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we have:

\[ \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta. \]

Thus, the integral becomes:

\[ \int \frac{\sec^2 \theta}{\tan^2 \theta \sec \theta} \, d\theta = \int \frac{\sec \theta}{\tan^2 \theta} \, d\theta. \]

### Step 4: Rewrite in terms of sine and cosine
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, we can rewrite the integral as:

\[ \int \frac{1/\cos \theta}{(\sin^2 \theta / \cos^2 \theta)} \, d\theta = \int \frac{\cos^2 \theta}{\sin^2 \theta} \, d\theta = \int \cot^2 \theta \, d\theta. \]

### Step 5: Use the identity for cotangent
We know that:

\[ \cot^2 \theta = \csc^2 \theta - 1. \]

Thus, we can rewrite the integral:

\[ \int \cot^2 \theta \, d\theta = \int (\csc^2 \theta - 1) \, d\theta = \int \csc^2 \theta \, d\theta - \int 1 \, d\theta. \]

### Step 6: Integrate
The integral of \( \csc^2 \theta \) is:

\[ -\cot \theta, \]

and the integral of \( 1 \) is:

\[ \theta. \]

So we have:

\[ \int \cot^2 \theta \, d\theta = -\cot \theta - \theta + C. \]

### Step 7: Substitute back to \( x \)
Recall that \( x = \tan \theta \), so:

\[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{x}. \]

Also, from the triangle formed by \( \tan \theta \), we have:

\[ \sin \theta = \frac{x}{\sqrt{1+x^2}}, \quad \cos \theta = \frac{1}{\sqrt{1+x^2}}. \]

Thus,

\[ \theta = \tan^{-1}(x). \]

Putting it all together, we have:

\[ -\cot \theta - \theta + C = -\frac{1}{x} - \tan^{-1}(x) + C. \]

### Final Answer
The evaluated integral is:

\[ \int \frac{1}{x^2 \sqrt{1+x^2}} \, dx = -\frac{1}{x} - \tan^{-1}(x) + C. \]