Prove that : `int (1)/(x[6(logx)^(2)+7logx+2]] dx= log |(1+log x^(2))/(2+log x^(3))|+c`

Answer

To prove that

\[ \int \frac{1}{x(6(\log x)^2 + 7\log x + 2)} \, dx = \log \left| \frac{1 + \log x^2}{2 + \log x^3} \right| + C, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, differentiating both sides gives us:

\[ \frac{1}{x} \, dx = dt. \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:

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

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

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

Multiplying through by the denominator \( 6t^2 + 7t + 2 \) gives:

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

### Step 4: Solve for A and B
Expanding the right-hand side:

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

Now, we can set up a system of equations by equating coefficients:

1. \( 3A + 2B = 0 \) (coefficient of \( t \))
2. \( 2A + B = 1 \) (constant term)

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

\[ B = -\frac{3}{2}A. \]

Substituting this into the second equation:

\[ 2A - \frac{3}{2}A = 1 \implies \frac{1}{2}A = 1 \implies A = 2. \]

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

\[ B = -\frac{3}{2}(2) = -3. \]

### Step 5: Rewrite the Integral
Now substituting \( A \) and \( B \) back into the partial fractions gives:

\[ \int \left( \frac{2}{2t + 1} - \frac{3}{3t + 2} \right) dt. \]

### Step 6: Integrate
Now we can integrate each term separately:

\[ \int \frac{2}{2t + 1} \, dt - \int \frac{3}{3t + 2} \, dt. \]

The integrals yield:

\[ 2 \log |2t + 1| - 3 \log |3t + 2| + C. \]

### Step 7: Combine the Logs
Using the properties of logarithms, we can combine these:

\[ \log |(2t + 1)^2| - \log |(3t + 2)^3| = \log \left| \frac{(2t + 1)^2}{(3t + 2)^3} \right| + C. \]

### Step 8: Substitute Back
Now, substituting back \( t = \log x \):

\[ 2t + 1 = 2\log x + 1 = \log x^2 + 1, \]
\[ 3t + 2 = 3\log x + 2 = \log x^3 + 2. \]

Thus, we have:

\[ \log \left| \frac{(1 + \log x^2)}{(2 + \log x^3)} \right| + C. \]

### Conclusion
Therefore, we have proved that:

\[ \int \frac{1}{x(6(\log x)^2 + 7\log x + 2)} \, dx = \log \left| \frac{1 + \log x^2}{2 + \log x^3} \right| + C. \]