f(x) is double differentiable function. Tangent at (1, f(1)) and (3, (f(3)) cuts positive x-axis at an angle `pi/6` and `pi/4` then value of `int_1^3(f'(x)^2+1)f''(x)dx` is:

Options

A. `4/3-10/(9sqrt3)`
B. `3/4-10/(9sqrt3)`
C. `4/3-10/(sqrt3)`
D. `4/3-1/(9sqrt3)`

Answer

To solve the problem, we need to analyze the tangents at the points (1, f(1)) and (3, f(3)) and then evaluate the integral given.

### Step 1: Determine the slopes of the tangents
The slopes of the tangents at the points (1, f(1)) and (3, f(3)) can be determined from the angles given.

- For the tangent at (1, f(1)), the angle with the positive x-axis is \(\frac{\pi}{6}\):
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \implies f'(1) = \frac{1}{\sqrt{3}} \]

- For the tangent at (3, f(3)), the angle with the positive x-axis is \(\frac{\pi}{4}\):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \implies f'(3) = 1 \]

### Step 2: Set up the integral
We need to evaluate the integral:
\[ I = \int_1^3 (f'(x)^2 + 1) f''(x) \, dx \]

### Step 3: Use substitution
Let \( t = f'(x) \). Then, by the chain rule, we have:
\[ dt = f''(x) \, dx \implies dx = \frac{dt}{f''(x)} \]

### Step 4: Change the limits of integration
When \( x = 1 \), \( f'(1) = \frac{1}{\sqrt{3}} \) and when \( x = 3 \), \( f'(3) = 1 \). Thus, the limits of integration change from \( x = 1 \) to \( x = 3 \) into \( t = \frac{1}{\sqrt{3}} \) to \( t = 1 \).

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

### Step 6: Evaluate the integral
Now we can compute the integral:
\[ I = \int_{\frac{1}{\sqrt{3}}}^{1} (t^2 + 1) \, dt = \int_{\frac{1}{\sqrt{3}}}^{1} t^2 \, dt + \int_{\frac{1}{\sqrt{3}}}^{1} 1 \, dt \]

Calculating each part:
1. For \( \int t^2 \, dt \):
\[ \int t^2 \, dt = \frac{t^3}{3} \Big|_{\frac{1}{\sqrt{3}}}^{1} = \frac{1^3}{3} - \frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3} = \frac{1}{3} - \frac{1}{9\sqrt{3}} \]

2. For \( \int 1 \, dt \):
\[ \int 1 \, dt = t \Big|_{\frac{1}{\sqrt{3}}}^{1} = 1 - \frac{1}{\sqrt{3}} \]

Combining these results:
\[ I = \left(\frac{1}{3} - \frac{1}{9\sqrt{3}}\right) + \left(1 - \frac{1}{\sqrt{3}}\right) \]
\[ = \frac{1}{3} + 1 - \frac{1}{\sqrt{3}} - \frac{1}{9\sqrt{3}} = \frac{4}{3} - \frac{4}{9\sqrt{3}} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{4}{3} - \frac{4}{9\sqrt{3}} \]