`int(sqrt(x^(2)+1))/(x^(4))dx` is equal to

Options

A. `(1)/(3)(1+(1)/(x^(2)))^(3//2)+C`
B. `(2)/(3)(1+(1)/(x^(2)))^(3//2)+C`
C. `-(1)/(3)(1+(1)/(x^(2)))^(3//2)+C`
D. `(3)/(2)(1+(1)/(x^(2)))^(3//2)+C`

Explanation

To solve the integral \( I = \int \frac{\sqrt{x^2 + 1}}{x^4} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, we have:
\[ dx = \sec^2 \theta \, d\theta \]
Also, we know that:
\[ \sqrt{x^2 + 1} = \sqrt{\tan^2 \theta + 1} = \sqrt{\sec^2 \theta} = \sec \theta \]
Now substituting these into the integral, we get:
\[ I = \int \frac{\sec \theta}{\tan^4 \theta} \sec^2 \theta \, d\theta \]
This simplifies to:
\[ I = \int \frac{\sec^3 \theta}{\tan^4 \theta} \, d\theta \]

### Step 2: Rewrite in terms of sine and cosine
Recall that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]
Thus, we can rewrite the integral:
\[ I = \int \frac{1}{\cos^3 \theta} \cdot \frac{\cos^4 \theta}{\sin^4 \theta} \, d\theta = \int \frac{\cos \theta}{\sin^4 \theta} \, d\theta \]

### Step 3: Substitution for sine
Now, let \( t = \sin \theta \). Then, \( dt = \cos \theta \, d\theta \). The integral becomes:
\[ I = \int \frac{1}{t^4} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ I = \int t^{-4} \, dt = \frac{t^{-3}}{-3} + C = -\frac{1}{3t^3} + C \]

### Step 5: Back substitute
Recall that \( t = \sin \theta \) and \( \theta = \tan^{-1}(x) \). Therefore:
\[ \sin \theta = \frac{x}{\sqrt{x^2 + 1}} \]
Thus,
\[ t^3 = \left(\frac{x}{\sqrt{x^2 + 1}}\right)^3 = \frac{x^3}{(x^2 + 1)^{3/2}} \]
Substituting back gives:
\[ I = -\frac{1}{3} \cdot \frac{(x^2 + 1)^{3/2}}{x^3} + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = -\frac{(x^2 + 1)^{3/2}}{3x^3} + C \]