`int(sqrt(x^2+1))/(x^4)dx=`

Options

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

Explanation

To solve the integral \( \int \frac{\sqrt{x^2 + 1}}{x^4} \, dx \), we can use a trigonometric substitution. Let's go through the solution step by step.

### Step 1: Trigonometric Substitution
Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{x^2 + 1} = \sqrt{\tan^2(\theta) + 1} = \sqrt{\sec^2(\theta)} = \sec(\theta) \]
Also, the differential \( dx \) is given by:
\[ dx = \sec^2(\theta) \, d\theta \]

### Step 2: Substitute in the Integral
Now, substituting \( x \) and \( dx \) into the integral, we get:
\[ \int \frac{\sqrt{x^2 + 1}}{x^4} \, dx = \int \frac{\sec(\theta)}{\tan^4(\theta)} \sec^2(\theta) \, d\theta \]
This simplifies to:
\[ \int \frac{\sec^3(\theta)}{\tan^4(\theta)} \, d\theta \]

### Step 3: Rewrite in Terms of Sine and Cosine
Using the identities \( \sec(\theta) = \frac{1}{\cos(\theta)} \) and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can rewrite the integral:
\[ \int \frac{1}{\cos^3(\theta)} \cdot \frac{\cos^4(\theta)}{\sin^4(\theta)} \, d\theta = \int \frac{\cos(\theta)}{\sin^4(\theta)} \, d\theta \]

### Step 4: Simplifying the Integral
This can be expressed as:
\[ \int \cos(\theta) \cdot \csc^4(\theta) \, d\theta \]
Now, we can use the substitution \( u = \sin(\theta) \), which gives \( du = \cos(\theta) \, d\theta \). The integral becomes:
\[ \int \frac{1}{u^4} \, du \]

### Step 5: Integrate
Now, we can integrate:
\[ \int u^{-4} \, du = -\frac{1}{3} u^{-3} + C = -\frac{1}{3 \sin^3(\theta)} + C \]

### Step 6: Back Substitute
Now, substituting back for \( \sin(\theta) \):
Since \( \sin(\theta) = \frac{x}{\sqrt{x^2 + 1}} \), we have:
\[ -\frac{1}{3} \left(\frac{x}{\sqrt{x^2 + 1}}\right)^{-3} + C = -\frac{1}{3} \cdot \frac{(\sqrt{x^2 + 1})^3}{x^3} + C \]

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