The value of `int_(1)^(e)(dx)/(6x(logx)^(2)+7x(logx)+2x)=`

Options

A. `log_(e)((15)/(2))`
B. `log_(e)((6)/(5))`
C. `log_(e)((3)/(10))`
D. `(1)/(5)log_(e)((8)/(3))`

Answer

To solve the integral

\[ I = \int_{1}^{e} \frac{dx}{6x(\log x)^{2} + 7x \log x + 2x} \]

we will follow these steps:

### Step 1: Factor out \( x \) from the denominator
We can factor out \( x \) from the denominator:

\[ I = \int_{1}^{e} \frac{dx}{x(6(\log x)^{2} + 7\log x + 2)} \]

### Step 2: Substitute \( t = \log x \)
Next, we will make the substitution \( t = \log x \). Then, \( dx = e^t dt \) and when \( x = 1 \), \( t = 0 \) and when \( x = e \), \( t = 1 \). The integral becomes:

\[ I = \int_{0}^{1} \frac{e^t dt}{e^t(6t^{2} + 7t + 2)} = \int_{0}^{1} \frac{dt}{6t^{2} + 7t + 2} \]

### Step 3: Factor the quadratic
Now we need to factor the quadratic in the denominator \( 6t^{2} + 7t + 2 \). We look for two numbers that multiply to \( 12 \) (the product of \( 6 \) and \( 2 \)) and add to \( 7 \). The numbers \( 3 \) and \( 4 \) work:

\[ 6t^{2} + 3t + 4t + 2 = 3t(2t + 1) + 2(2t + 1) = (2t + 1)(3t + 2) \]

So we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{dt}{(2t + 1)(3t + 2)} \]

### Step 4: Partial fraction decomposition
Next, we will use partial fractions to decompose the integrand:

\[ \frac{1}{(2t + 1)(3t + 2)} = \frac{A}{2t + 1} + \frac{B}{3t + 2} \]

Multiplying through by the denominator \( (2t + 1)(3t + 2) \) gives:

\[ 1 = A(3t + 2) + B(2t + 1) \]

Expanding and rearranging:

\[ 1 = (3A + 2B)t + (2A + B) \]

Setting up the system of equations:

1. \( 3A + 2B = 0 \)
2. \( 2A + B = 1 \)

From the second equation, we can express \( B \) in terms of \( A \):

\[ B = 1 - 2A \]

Substituting into the first equation:

\[ 3A + 2(1 - 2A) = 0 \implies 3A + 2 - 4A = 0 \implies -A + 2 = 0 \implies A = 2 \]

Substituting \( A = 2 \) back into the equation for \( B \):

\[ B = 1 - 2(2) = 1 - 4 = -3 \]

Thus, we have:

\[ \frac{1}{(2t + 1)(3t + 2)} = \frac{2}{2t + 1} - \frac{3}{3t + 2} \]

### Step 5: Integrate
Now we can integrate:

\[ I = \int_{0}^{1} \left( \frac{2}{2t + 1} - \frac{3}{3t + 2} \right) dt \]

Calculating the integrals separately:

1. \(\int \frac{2}{2t + 1} dt = 2 \cdot \frac{1}{2} \ln |2t + 1| = \ln |2t + 1|\)
2. \(\int \frac{3}{3t + 2} dt = 3 \cdot \frac{1}{3} \ln |3t + 2| = \ln |3t + 2|\)

Thus,

\[ I = \left[ \ln(2t + 1) - \ln(3t + 2) \right]_{0}^{1} \]

### Step 6: Evaluate the limits
Evaluating at the limits:

\[ I = \left( \ln(2 \cdot 1 + 1) - \ln(3 \cdot 1 + 2) \right) - \left( \ln(2 \cdot 0 + 1) - \ln(3 \cdot 0 + 2) \right) \]

This simplifies to:

\[ I = \left( \ln(3) - \ln(5) \right) - \left( \ln(1) - \ln(2) \right) = \ln(3) - \ln(5) + \ln(2) \]

Using properties of logarithms:

\[ I = \ln\left(\frac{3 \cdot 2}{5}\right) = \ln\left(\frac{6}{5}\right) \]

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

\[ \boxed{\ln\left(\frac{6}{5}\right)} \]