The tangent, represented by the graph of the function `y=f(x),` at the point with abscissa x = 1 form an angle of `pi//6`, at the point x = 2 form an angle of `pi//3` and at the point x = 3 form and angle of `pi//4`. Then, find the value of,
`int_(1)^(3)f'(x)f''(x)dx+int_(2)^(3)f''(x)dx.`

Answer

To solve the given problem, we need to find the value of the expression:

\[ I = \int_{1}^{3} f'(x) f''(x) \, dx + \int_{2}^{3} f''(x) \, dx \]

### Step 1: Find the values of \( f'(x) \)

We know that the slope of the tangent line at a point on the curve \( y = f(x) \) is given by \( f'(x) = \tan(\theta) \), where \( \theta \) is the angle the tangent makes with the positive x-axis.

1. At \( x = 1 \), the angle is \( \frac{\pi}{6} \):
\[ f'(1) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

2. At \( x = 2 \), the angle is \( \frac{\pi}{3} \):
\[ f'(2) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

3. At \( x = 3 \), the angle is \( \frac{\pi}{4} \):
\[ f'(3) = \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 2: Evaluate the first integral \( \int_{1}^{3} f'(x) f''(x) \, dx \)

Using integration by parts, we can express the integral as follows:

\[ \int f'(x) f''(x) \, dx = \frac{1}{2} (f'(x))^2 + C \]

Thus, we have:

\[ \int_{1}^{3} f'(x) f''(x) \, dx = \left[ \frac{1}{2} (f'(x))^2 \right]_{1}^{3} \]

Calculating this:

\[ = \frac{1}{2} (f'(3))^2 - \frac{1}{2} (f'(1))^2 \]

Substituting the values we found:

\[ = \frac{1}{2} (1)^2 - \frac{1}{2} \left(\frac{1}{\sqrt{3}}\right)^2 \]
\[ = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{3} \]
\[ = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

Let \( I_1 = \frac{1}{3} \).

### Step 3: Evaluate the second integral \( \int_{2}^{3} f''(x) \, dx \)

Integrating \( f''(x) \):

\[ \int_{2}^{3} f''(x) \, dx = \left[ f'(x) \right]_{2}^{3} = f'(3) - f'(2) \]

Substituting the values:

\[ = 1 - \sqrt{3} \]

Let \( I_2 = 1 - \sqrt{3} \).

### Step 4: Combine the results

Now we can find the total value of \( I \):

\[ I = I_1 + I_2 = \frac{1}{3} + (1 - \sqrt{3}) \]
\[ = \frac{1}{3} + 1 - \sqrt{3} = \frac{1}{3} + \frac{3}{3} - \sqrt{3} = \frac{4}{3} - \sqrt{3} \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{\frac{4}{3} - \sqrt{3}} \]