`int_(1)^(e )(1)/(6x(log x)^(2)+7x log x + 2x)dx=`

Options

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

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the Integral
We start by factoring out \(x\) from the denominator:

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

### Step 2: Substitute \(t = \log x\)
We can use the substitution \(t = \log x\). Therefore, \(x = e^t\) and \(dx = e^t \, dt\). The limits change as follows:
- When \(x = 1\), \(t = \log 1 = 0\).
- When \(x = e\), \(t = \log e = 1\).

Thus, the integral becomes:

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

### Step 3: Partial Fraction Decomposition
Next, we need to perform partial fraction decomposition on the integrand:

\[ \frac{1}{6t^2 + 7t + 2}. \]

We can express this as:

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

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

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

### Step 4: Solve for \(A\) and \(B\)
To find \(A\) and \(B\), we can substitute convenient values for \(t\):

1. Let \(t = -\frac{1}{2}\):

\[ 1 = A(0) + B(3(-\frac{1}{2}) + 2) \Rightarrow 1 = B(0.5) \Rightarrow B = 2. \]

2. Let \(t = -\frac{2}{3}\):

\[ 1 = A(2(-\frac{2}{3}) + 1) + B(0) \Rightarrow 1 = A(-\frac{4}{3} + 1) \Rightarrow 1 = A(-\frac{1}{3}) \Rightarrow A = -3. \]

Thus, we have:

\[ \frac{1}{6t^2 + 7t + 2} = \frac{-3}{3t + 2} + \frac{2}{2t + 1}. \]

### Step 5: Integrate Each Term
Now we can integrate each term separately:

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

Calculating these integrals:

1. For \(-3 \int \frac{1}{3t + 2} dt\):

\[ -3 \cdot \frac{1}{3} \log |3t + 2| \bigg|_{0}^{1} = -\log(5) + \log(2) = \log\left(\frac{2}{5}\right). \]

2. For \(2 \int \frac{1}{2t + 1} dt\):

\[ 2 \cdot \frac{1}{2} \log |2t + 1| \bigg|_{0}^{1} = \log(3) - \log(1) = \log(3). \]

### Step 6: Combine the Results
Combining the results from both integrals:

\[ \log\left(\frac{2}{5}\right) + \log(3) = \log\left(\frac{6}{5}\right). \]

### Final Answer
Thus, the final result of the integral is:

\[ \int_{1}^{e} \frac{1}{6x (\log x)^2 + 7x \log x + 2x} \, dx = \log\left(\frac{6}{5}\right). \]