The value of `int_0^1 (x^(3))/(1+x^(8))dx` is

Options

A. `(pi)/(4)`
B. `(pi)/(8)`
C. `(pi)/(16)`
D. none of these

Answer

To solve the integral \( I = \int_0^1 \frac{x^3}{1+x^8} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Define the integral
Let:
\[ I = \int_0^1 \frac{x^3}{1+x^8} \, dx \]

### Step 2: Use substitution
We will use the substitution \( t = x^4 \). Then, we differentiate:
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]
Also, when \( x = 0 \), \( t = 0^4 = 0 \) and when \( x = 1 \), \( t = 1^4 = 1 \).

### Step 3: Substitute in the integral
Now, we can express \( x^3 \) in terms of \( t \):
\[ x = t^{1/4} \quad \Rightarrow \quad x^3 = (t^{1/4})^3 = t^{3/4} \]
Thus, the integral becomes:
\[ I = \int_0^1 \frac{t^{3/4}}{1 + (t^{1/4})^8} \cdot \frac{dt}{4t^{3/4}} = \frac{1}{4} \int_0^1 \frac{1}{1+t^2} \, dt \]

### Step 4: Simplify the integral
Now we can simplify the integral:
\[ I = \frac{1}{4} \int_0^1 \frac{1}{1+t^2} \, dt \]

### Step 5: Evaluate the integral
The integral \( \int \frac{1}{1+t^2} \, dt \) is known to be:
\[ \int \frac{1}{1+t^2} \, dt = \tan^{-1}(t) \]
Thus, we evaluate:
\[ \int_0^1 \frac{1}{1+t^2} \, dt = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 6: Final calculation
Now substituting back:
\[ I = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16} \]

### Conclusion
The value of the integral is:
\[ \boxed{\frac{\pi}{16}} \]