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 and properties of definite integrals. Let's go through the solution step by step.

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

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

### Step 3: Rewrite the integral
Substituting into the integral, we have:
\[ I = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} \frac{-\frac{1}{t^2}}{1 + \left(\frac{1}{t}\right)^2 + \left(\frac{1}{t}\right)^3 + \left(\frac{1}{t}\right)^5} dt \]
This simplifies to:
\[ I = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} \frac{-\frac{1}{t^2}}{\frac{1 + t^2 + t^3 + t^5}{t^5}} dt = \int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}} \frac{-t^3}{1 + t^2 + t^3 + t^5} dt \]

### Step 4: Change the limits back
Now, we can change the limits back:
\[ I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{t^3}{1 + t^2 + t^3 + t^5} dt \]

### Step 5: 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 equations gives:
\[ 2I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1 + x^3}{1 + x^2 + x^3 + x^5} dx \]

### Step 6: Simplify the integral
The denominator can be factored:
\[ 1 + x^2 + x^3 + x^5 = (1 + x^3) + (x^2 + x^5) \]
So we can write:
\[ 2I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1 + x^3}{1 + x^2 + x^3 + x^5} dx = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1 + x^2 + x^3 + x^5} dx + \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{x^3}{1 + x^2 + x^3 + x^5} dx \]

### Step 7: Evaluate the integral
The integral \( \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{1 + x^2} dx \) can be computed:
\[ \int \frac{1}{1 + x^2} dx = \tan^{-1}(x) \]
Thus:
\[ I = \frac{1}{2} \left[ \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \right] \]
Using the values \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \) and \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \):
\[ 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]
Thus:
\[ I = \frac{\pi}{12} \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{\pi}{12}} \]