The solution of differential equation `(dy)/(dx)+(2xy)/(1+x^(2))=(1)/(1+x^(2))^(2)` is

Options

A. `y(1+x^(2))=C+tan^(-1)x`
B. `(y)/(1+x^(2))=C+tan^(-1)x`
C. `y log (1+x^(2))=C+tan^(-1)x`
D. `(1+x^(2))=C+sin^(-1)x`

Explanation

To solve the differential equation

\[ \frac{dy}{dx} + \frac{2xy}{1+x^2} = \frac{1}{(1+x^2)^2}, \]

we will follow these steps:

### Step 1: Identify the components of the linear differential equation
This equation is of the form

\[ \frac{dy}{dx} + P(x)y = Q(x), \]

where

\[ P(x) = \frac{2x}{1+x^2} \quad \text{and} \quad Q(x) = \frac{1}{(1+x^2)^2}. \]

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2x}{1+x^2} \, dx}. \]

To compute the integral:

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

we can use the substitution \( t = 1 + x^2 \), which gives \( dt = 2x \, dx \). Thus, the integral becomes:

\[ \int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \ln |t| + C = \ln(1+x^2) + C. \]

So, the integrating factor is:

\[ \mu(x) = e^{\ln(1+x^2)} = 1+x^2. \]

### Step 3: Multiply the entire differential equation by the integrating factor
Now we multiply the entire differential equation by \( 1+x^2 \):

\[ (1+x^2)\frac{dy}{dx} + (1+x^2)\frac{2xy}{1+x^2} = (1+x^2)\frac{1}{(1+x^2)^2}. \]

This simplifies to:

\[ (1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}. \]

### Step 4: Rewrite the left-hand side
The left-hand side can be rewritten as:

\[ \frac{d}{dx}(y(1+x^2)). \]

Thus, we have:

\[ \frac{d}{dx}(y(1+x^2)) = \frac{1}{1+x^2}. \]

### Step 5: Integrate both sides
Now we integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}(y(1+x^2)) \, dx = \int \frac{1}{1+x^2} \, dx. \]

The left side simplifies to:

\[ y(1+x^2) = \tan^{-1}(x) + C, \]

where \( C \) is the constant of integration.

### Step 6: Solve for \( y \)
Finally, we solve for \( y \):

\[ y = \frac{\tan^{-1}(x) + C}{1+x^2}. \]

### Final Solution
Thus, the solution of the differential equation is:

\[ y = \frac{\tan^{-1}(x) + C}{1+x^2}. \]