The value of the integral `I=int_((1)/(sqrt3))^(sqrt3)(dx)/(1+x^(2)+x^(3)+x^(5))` is equal to

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(12)`
D. `(pi)/(6)`

Answer

To solve the integral

\[ I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1 + x^2 + x^3 + x^5}, \]

we will use a substitution method and properties of definite integrals. Here are the steps:

### Step 1: Substitution
Let \( t = \frac{1}{x} \). Then, we have \( x = \frac{1}{t} \) and \( dx = -\frac{1}{t^2} dt \).

### Step 2: Change the limits
When \( x = \frac{1}{\sqrt{3}} \), \( t = \sqrt{3} \) and when \( x = \sqrt{3} \), \( t = \frac{1}{\sqrt{3}} \). Thus, the limits of integration change from \( \frac{1}{\sqrt{3}} \) to \( \sqrt{3} \) into \( \sqrt{3} \) to \( \frac{1}{\sqrt{3}} \).

### Step 3: Rewrite the integral
Now, substituting into the integral, we get:

\[ I = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} \frac{-\frac{1}{t^2} dt}{1 + \left(\frac{1}{t}\right)^2 + \left(\frac{1}{t}\right)^3 + \left(\frac{1}{t}\right)^5}. \]

### Step 4: Simplify the denominator
The denominator becomes:

\[ 1 + \frac{1}{t^2} + \frac{1}{t^3} + \frac{1}{t^5} = \frac{t^5 + t^3 + t^2 + 1}{t^5}. \]

Thus, we can rewrite the integral as:

\[ I = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} -\frac{t^5}{t^2(t^5 + t^3 + t^2 + 1)} dt = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} -\frac{t^3}{t^5 + t^3 + t^2 + 1} dt. \]

### Step 5: Change the limits back
Changing the limits back gives:

\[ I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{t^3}{t^5 + t^3 + t^2 + 1} dt. \]

### Step 6: Combine the integrals
Now we have two expressions for \( I \):

1. \( I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1 + x^2 + x^3 + x^5} \)
2. \( I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{x^3}{1 + x^2 + x^3 + x^5} dx \)

Adding these two integrals gives:

\[ 2I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1 + x^3}{1 + x^2 + x^3 + x^5} dx. \]

### Step 7: Simplify the denominator
The denominator can be factored as:

\[ 1 + x^2 + x^3 + x^5 = (1 + x^3)(1 + x^2). \]

Thus,

\[ 2I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1 + x^3}{(1 + x^3)(1 + x^2)} dx = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1 + x^2} dx. \]

### Step 8: Evaluate the integral
The integral \( \int \frac{1}{1 + x^2} dx \) is \( \tan^{-1}(x) \). Therefore:

\[ 2I = \left[ \tan^{-1}(x) \right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \]

### Step 9: Calculate the values
We know that:

\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \quad \text{and} \quad \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. \]

Thus,

\[ 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. \]

### Step 10: Solve for \( I \)
Finally, we find \( I \):

\[ I = \frac{\pi}{12}. \]

### Final Answer
The value of the integral is

\[ \boxed{\frac{\pi}{12}}. \]