The integral `16int_(1)^(2)(dx)/(x^(3)(x^(2)+2)^(2))` is equal to

Options

A. `(11)/(6)-log_(e)4`
B. `(11)/(12)+log_(e)4`
C. `(11)/(6)+log_(e)4`
D. `(11)/(12)-log_(e)4`

Answer

To solve the integral

\[ 16 \int_{1}^{2} \frac{dx}{x^3 (x^2 + 2)^2} \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = 16 \int_{1}^{2} \frac{dx}{x^3 (x^2 + 2)^2} \]

### Step 2: Substitution
Let \( t = x^2 + 2 \). Then, we differentiate to find \( dt \):

\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]

Now, we also need to change the limits of integration. When \( x = 1 \):

\[ t = 1^2 + 2 = 3 \]

When \( x = 2 \):

\[ t = 2^2 + 2 = 6 \]

### Step 3: Express \( x \) in terms of \( t \)
From \( t = x^2 + 2 \), we can express \( x^2 \) as:

\[ x^2 = t - 2 \quad \Rightarrow \quad x = \sqrt{t - 2} \]

### Step 4: Substitute into the Integral
Now we substitute everything back into the integral:

\[ I = 16 \int_{3}^{6} \frac{1}{(\sqrt{t - 2})^3 (t)^2} \cdot \frac{dt}{2\sqrt{t - 2}} \]

This simplifies to:

\[ I = 16 \int_{3}^{6} \frac{1}{2(t - 2)^{2} t^2} \, dt \]

### Step 5: Simplify the Integral
Now we can factor out the constants:

\[ I = 8 \int_{3}^{6} \frac{1}{(t - 2)^{2} t^2} \, dt \]

### Step 6: Split the Integral
We can split the integral into two parts:

\[ I = 8 \left( \int_{3}^{6} \frac{1}{(t - 2)^{2}} \, dt - \int_{3}^{6} \frac{1}{t^2} \, dt \right) \]

### Step 7: Evaluate the Integrals
1. For the first integral:

\[ \int \frac{1}{(t - 2)^{2}} \, dt = -\frac{1}{t - 2} \]

Evaluating from 3 to 6 gives:

\[ -\left[ \frac{1}{6 - 2} - \frac{1}{3 - 2} \right] = -\left[ \frac{1}{4} - 1 \right] = -\left[ \frac{1}{4} - \frac{4}{4} \right] = -\left[ -\frac{3}{4} \right] = \frac{3}{4} \]

2. For the second integral:

\[ \int \frac{1}{t^2} \, dt = -\frac{1}{t} \]

Evaluating from 3 to 6 gives:

\[ -\left[ \frac{1}{6} - \frac{1}{3} \right] = -\left[ \frac{1}{6} - \frac{2}{6} \right] = -\left[ -\frac{1}{6} \right] = \frac{1}{6} \]

### Step 8: Combine the Results
Now, we combine the results:

\[ I = 8 \left( \frac{3}{4} - \frac{1}{6} \right) \]

Finding a common denominator (12):

\[ I = 8 \left( \frac{9}{12} - \frac{2}{12} \right) = 8 \cdot \frac{7}{12} = \frac{56}{12} = \frac{14}{3} \]

### Final Answer
Thus, the value of the integral is:

\[ \frac{14}{3} \]