Evaluate `int(tan^(-1)x)/(x^(4))dx`.

Answer

To evaluate the integral \( I = \int \frac{\tan^{-1}(x)}{x^4} \, dx \), we will use integration by parts. Let's go through the solution step by step.

### Step 1: Identify \( u \) and \( dv \)
We choose:
- \( u = \tan^{-1}(x) \) (which we will differentiate)
- \( dv = \frac{1}{x^4} \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)
Now we compute \( du \) and \( v \):
- \( du = \frac{1}{1 + x^2} \, dx \)
- To find \( v \), we integrate \( dv \):
\[ v = \int \frac{1}{x^4} \, dx = -\frac{1}{3x^3} \]

### Step 3: Apply Integration by Parts Formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = \tan^{-1}(x) \left(-\frac{1}{3x^3}\right) - \int \left(-\frac{1}{3x^3}\right) \left(\frac{1}{1 + x^2}\right) \, dx \]
This simplifies to:
\[ I = -\frac{\tan^{-1}(x)}{3x^3} + \frac{1}{3} \int \frac{1}{x^3(1 + x^2)} \, dx \]

### Step 4: Evaluate the Remaining Integral
Now we need to evaluate:
\[ \int \frac{1}{x^3(1 + x^2)} \, dx \]
We can use partial fractions to simplify this integral:
\[ \frac{1}{x^3(1 + x^2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{1 + x^2} \]
Multiplying through by \( x^3(1 + x^2) \) and equating coefficients will yield values for \( A, B, C, \) and \( D \).

### Step 5: Solve for Coefficients
After setting up the equation and solving for the coefficients, we find:
- \( A = 0 \)
- \( B = 0 \)
- \( C = 1 \)
- \( D = -1 \)

Thus, we can rewrite the integral as:
\[ \int \left( \frac{1}{x^3} - \frac{1}{1 + x^2} \right) \, dx \]

### Step 6: Integrate Each Term
Now we integrate each term:
1. \( \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} \)
2. \( \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) \)

Putting it all together, we have:
\[ \int \frac{1}{x^3(1 + x^2)} \, dx = -\frac{1}{2x^2} - \tan^{-1}(x) \]

### Step 7: Substitute Back
Now substituting back into our expression for \( I \):
\[ I = -\frac{\tan^{-1}(x)}{3x^3} + \frac{1}{3} \left(-\frac{1}{2x^2} - \tan^{-1}(x)\right) \]

### Step 8: Simplify
Combining the terms, we get:
\[ I = -\frac{\tan^{-1}(x)}{3x^3} - \frac{1}{6x^2} - \frac{1}{3}\tan^{-1}(x) + C \]
Where \( C \) is the constant of integration.

### Final Answer
Thus, the final answer is:
\[ I = -\frac{\tan^{-1}(x)}{3x^3} - \frac{1}{6x^2} - \frac{1}{3}\tan^{-1}(x) + C \]