`int_(1)^(5) (sqrt""x)/(sqrt""(6-x) + sqrt""x) dx`=

Options

A. 1
B. `3//2`
C. 2
D. `5//2`

Answer

To solve the integral

\[ I = \int_{1}^{5} \frac{\sqrt{x}}{\sqrt{6-x} + \sqrt{x}} \, dx, \]

we can utilize a property of definite integrals. Specifically, we can use the property that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In this case, \(a = 1\) and \(b = 5\), so \(a + b = 6\). Thus, we can rewrite the integral as follows:

1. **Step 1: Rewrite the integral using the property.**

We replace \(x\) with \(6 - x\):

\[ I = \int_{1}^{5} \frac{\sqrt{6-x}}{\sqrt{x} + \sqrt{6-x}} \, dx. \]

2. **Step 2: Add the two integrals.**

Now, we have two expressions for \(I\):

\[ I = \int_{1}^{5} \frac{\sqrt{x}}{\sqrt{6-x} + \sqrt{x}} \, dx, \]
\[ I = \int_{1}^{5} \frac{\sqrt{6-x}}{\sqrt{x} + \sqrt{6-x}} \, dx. \]

Adding these two equations gives:

\[ 2I = \int_{1}^{5} \left( \frac{\sqrt{x}}{\sqrt{6-x} + \sqrt{x}} + \frac{\sqrt{6-x}}{\sqrt{x} + \sqrt{6-x}} \right) \, dx. \]

3. **Step 3: Simplify the integrand.**

Notice that the denominators are the same, so we can combine the numerators:

\[ 2I = \int_{1}^{5} \frac{\sqrt{x} + \sqrt{6-x}}{\sqrt{6-x} + \sqrt{x}} \, dx. \]

This simplifies to:

\[ 2I = \int_{1}^{5} 1 \, dx. \]

4. **Step 4: Evaluate the integral.**

The integral of 1 from 1 to 5 is simply the length of the interval:

\[ \int_{1}^{5} 1 \, dx = 5 - 1 = 4. \]

Therefore, we have:

\[ 2I = 4. \]

5. **Step 5: Solve for \(I\).**

Dividing both sides by 2 gives:

\[ I = \frac{4}{2} = 2. \]

Thus, the value of the integral is

\[ \boxed{2}. \]