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

Options

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

Explanation

To solve the integral \( \int \frac{1}{(x^2 + 1)^2} \, dx \), we will use the substitution method. Let's go through the steps:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, we know that:
\[ dx = \sec^2 \theta \, d\theta \]
Also, recall that:
\[ x^2 + 1 = \tan^2 \theta + 1 = \sec^2 \theta \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ \int \frac{1}{(x^2 + 1)^2} \, dx = \int \frac{1}{(\sec^2 \theta)^2} \sec^2 \theta \, d\theta \]
This simplifies to:
\[ \int \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta = \int \cos^2 \theta \, d\theta \]

### Step 3: Use the Identity for Cosine
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \), we can rewrite the integral:
\[ \int \cos^2 \theta \, d\theta = \int \frac{1 + \cos(2\theta)}{2} \, d\theta \]
This can be separated into two integrals:
\[ = \frac{1}{2} \int 1 \, d\theta + \frac{1}{2} \int \cos(2\theta) \, d\theta \]

### Step 4: Integrate
Now we can integrate each part:
\[ \frac{1}{2} \int 1 \, d\theta = \frac{\theta}{2} \]
\[ \frac{1}{2} \int \cos(2\theta) \, d\theta = \frac{1}{2} \cdot \frac{\sin(2\theta)}{2} = \frac{\sin(2\theta)}{4} \]
Putting it all together, we have:
\[ \int \cos^2 \theta \, d\theta = \frac{\theta}{2} + \frac{\sin(2\theta)}{4} + C \]

### Step 5: Back Substitute
Now we need to substitute back in terms of \( x \). Recall that \( \theta = \tan^{-1}(x) \), so:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
From the right triangle formed by \( \tan \theta = \frac{x}{1} \):
- Opposite side (p) = \( x \)
- Adjacent side (b) = \( 1 \)
- Hypotenuse (h) = \( \sqrt{x^2 + 1} \)

Thus:
\[ \sin(\theta) = \frac{x}{\sqrt{x^2 + 1}}, \quad \cos(\theta) = \frac{1}{\sqrt{x^2 + 1}} \]
Therefore:
\[ \sin(2\theta) = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} \]

### Final Result
Substituting back, we get:
\[ \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{1}{2} \tan^{-1}(x) + \frac{1}{4} \cdot \frac{2x}{x^2 + 1} + C \]
This simplifies to:
\[ \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{1}{2} \tan^{-1}(x) + \frac{x}{2(x^2 + 1)} + C \]