If `int_((1)/(3))^(3)|log_(e)x|dx=(m)/(n)log_(e)(n^2/e),` where `m` and `n` are co-prime natural numbers ,then `m^(2)+n^(2)-5` is equal to

Answer

To solve the integral \( \int_{\frac{1}{3}}^{3} |\log_e x| \, dx \) and express it in the form \( \frac{m}{n} \log_e \left( \frac{n^2}{e} \right) \), where \( m \) and \( n \) are co-prime natural numbers, we can follow these steps:

### Step 1: Split the Integral
The absolute value function \( |\log_e x| \) changes its behavior at \( x = 1 \). Therefore, we can split the integral at this point:

\[ \int_{\frac{1}{3}}^{3} |\log_e x| \, dx = \int_{\frac{1}{3}}^{1} -\log_e x \, dx + \int_{1}^{3} \log_e x \, dx \]

### Step 2: Evaluate the First Integral
For the first integral \( \int_{\frac{1}{3}}^{1} -\log_e x \, dx \):

Using integration by parts, let:
- \( u = \log_e x \) ⇒ \( du = \frac{1}{x} \, dx \)
- \( dv = dx \) ⇒ \( v = x \)

Then, the integration by parts formula \( \int u \, dv = uv - \int v \, du \) gives us:

\[ \int -\log_e x \, dx = -x \log_e x + \int x \cdot \frac{1}{x} \, dx = -x \log_e x + x \]

Now, we evaluate from \( \frac{1}{3} \) to \( 1 \):

\[ \left[-x \log_e x + x\right]_{\frac{1}{3}}^{1} = \left[-1 \cdot \log_e 1 + 1\right] - \left[-\frac{1}{3} \log_e \frac{1}{3} + \frac{1}{3}\right] \]

Since \( \log_e 1 = 0 \):

\[ = 1 - \left[-\frac{1}{3} \cdot (-\log_e 3) + \frac{1}{3}\right] = 1 - \left[\frac{1}{3} \log_e 3 + \frac{1}{3}\right] \]

\[ = 1 - \frac{1}{3} \log_e 3 - \frac{1}{3} = \frac{2}{3} - \frac{1}{3} \log_e 3 \]

### Step 3: Evaluate the Second Integral
Now consider the second integral \( \int_{1}^{3} \log_e x \, dx \):

Using integration by parts again:

\[ \int \log_e x \, dx = x \log_e x - x \]

Evaluating from \( 1 \) to \( 3 \):

\[ \left[x \log_e x - x\right]_{1}^{3} = \left[3 \log_e 3 - 3\right] - \left[1 \cdot \log_e 1 - 1\right] \]

Since \( \log_e 1 = 0 \):

\[ = (3 \log_e 3 - 3) - (0 - 1) = 3 \log_e 3 - 3 + 1 = 3 \log_e 3 - 2 \]

### Step 4: Combine the Results
Now, we combine both integrals:

\[ \int_{\frac{1}{3}}^{3} |\log_e x| \, dx = \left(\frac{2}{3} - \frac{1}{3} \log_e 3\right) + \left(3 \log_e 3 - 2\right) \]

Combining these:

\[ = \frac{2}{3} - \frac{1}{3} \log_e 3 + 3 \log_e 3 - 2 = \frac{2}{3} - 2 + \left(3 - \frac{1}{3}\right) \log_e 3 \]

\[ = \frac{2}{3} - \frac{6}{3} + \frac{8}{3} \log_e 3 = -\frac{4}{3} + \frac{8}{3} \log_e 3 \]

### Step 5: Express in Required Form
We want to express this in the form \( \frac{m}{n} \log_e \left( \frac{n^2}{e} \right) \):

\[ -\frac{4}{3} + \frac{8}{3} \log_e 3 = \frac{8}{3} \log_e 3 - \frac{4}{3} \]

This can be rewritten as:

\[ \frac{8}{3} \left( \log_e 3 - \frac{1}{2} \right) = \frac{8}{3} \log_e \left( \frac{3^2}{e} \right) \]

From this, we can identify \( m = 8 \) and \( n = 3 \).

### Step 6: Calculate \( m^2 + n^2 - 5 \)
Now we calculate \( m^2 + n^2 - 5 \):

\[ m^2 + n^2 - 5 = 8^2 + 3^2 - 5 = 64 + 9 - 5 = 68 \]

Thus, the final answer is:

\[ \boxed{68} \]