Evaluate the following Integrals :
`int(tan^(-1)x)/(x^(2))dx`

Answer

To evaluate the integral \( \int \frac{\tan^{-1} x}{x^2} \, dx \), we can use integration by parts. Let's denote:

- \( u = \tan^{-1} x \)
- \( dv = \frac{1}{x^2} \, dx \)

### Step 1: Differentiate \( u \) and Integrate \( dv \)

Now, we need to find \( du \) and \( v \):

1. Differentiate \( u \):
\[ du = \frac{1}{1 + x^2} \, dx \]

2. Integrate \( dv \):
\[ v = \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]

### Step 2: Apply Integration by Parts Formula

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute \( u \), \( v \), \( du \), and \( dv \):
\[ \int \frac{\tan^{-1} x}{x^2} \, dx = \left(-\frac{\tan^{-1} x}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{1 + x^2}\right) \, dx \]
This simplifies to:
\[ -\frac{\tan^{-1} x}{x} + \int \frac{1}{x(1 + x^2)} \, dx \]

### Step 3: Simplify the Remaining Integral

Now we need to evaluate the integral \( \int \frac{1}{x(1 + x^2)} \, dx \). We can use partial fraction decomposition:
\[ \frac{1}{x(1 + x^2)} = \frac{A}{x} + \frac{Bx + C}{1 + x^2} \]
Multiplying through by \( x(1 + x^2) \) gives:
\[ 1 = A(1 + x^2) + (Bx + C)x \]
Expanding and equating coefficients, we find:
- For \( x^2 \): \( A + B = 0 \)
- For \( x \): \( C = 0 \)
- Constant term: \( A = 1 \)

From \( A + B = 0 \), we have \( B = -1 \). Thus:
\[ \frac{1}{x(1 + x^2)} = \frac{1}{x} - \frac{x}{1 + x^2} \]

### Step 4: Integrate Each Term

Now we can integrate each term:
\[ \int \frac{1}{x(1 + x^2)} \, dx = \int \frac{1}{x} \, dx - \int \frac{x}{1 + x^2} \, dx \]
1. The first integral is:
\[ \int \frac{1}{x} \, dx = \ln |x| \]
2. The second integral can be solved using substitution \( t = 1 + x^2 \), \( dt = 2x \, dx \):
\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln |1 + x^2| \]

### Step 5: Combine Results

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

### Final Answer

Thus, the final result is:
\[ \int \frac{\tan^{-1} x}{x^2} \, dx = -\frac{\tan^{-1} x}{x} + \ln |x| - \frac{1}{2} \ln |1 + x^2| + C \]