`int ( x )/( 4+ x^(4)) dx` is equal to

Options

A. `(1)/( 4) tan^(-1) x^(2) + C`
B. `(1)/( 2) tan^(-1) ""(x^(2)/(2)) +C`
C. `(1)/(4) tan^(-1) ( (x^(2))/( 2)) + C`
D. `(1)/(2) tan^(-1) 2 x^(2) + C`

Answer

To solve the integral \(\int \frac{x}{4 + x^4} \, dx\), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{x}{4 + x^4} \, dx \]
Notice that \(x^4\) can be rewritten as \((x^2)^2\). Thus, we can express the denominator as:
\[ 4 + x^4 = 4 + (x^2)^2 \]

### Step 2: Use Substitution
Let’s use the substitution \(u = x^2\). Then, the differential \(du\) is given by:
\[ du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x} \]
Now, substituting \(x = \sqrt{u}\) gives us:
\[ dx = \frac{du}{2\sqrt{u}} \]

### Step 3: Substitute in the Integral
Substituting \(u\) into the integral:
\[ I = \int \frac{\sqrt{u}}{4 + u^2} \cdot \frac{du}{2\sqrt{u}} = \int \frac{1}{2(4 + u^2)} \, du \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{1}{4 + u^2} \, du \]

### Step 4: Integrate
The integral \(\int \frac{1}{4 + u^2} \, du\) can be solved using the formula:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
Here, \(a^2 = 4\) implies \(a = 2\). Thus, we have:
\[ \int \frac{1}{4 + u^2} \, du = \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + C \]

### Step 5: Substitute Back
Now substituting back \(u = x^2\):
\[ I = \frac{1}{2} \left( \frac{1}{2} \tan^{-1}\left(\frac{x^2}{2}\right) + C \right) = \frac{1}{4} \tan^{-1}\left(\frac{x^2}{2}\right) + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x}{4 + x^4} \, dx = \frac{1}{4} \tan^{-1}\left(\frac{x^2}{2}\right) + C \]

---