integral of (4x^2-3)/(x + 6)(x-3)(x-
8)) dx.

Answer:

To integrate the given expression, you need to perform partial fraction decomposition first.

The expression can be written as:

\[ \frac{4x^2 - 3}{(x + 6)(x - 3)(x - 8)} = \frac{A}{x + 6} + \frac{B}{x - 3} + \frac{C}{x - 8} \]

Now, let's find the values of \( A \), \( B \), and \( C \) by finding a common denominator:

\[ 4x^2 - 3 = A(x - 3)(x - 8) + B(x + 6)(x - 8) + C(x + 6)(x - 3) \]

Expand and equate coefficients:

\[ 4x^2 - 3 = A(x^2 - 11x + 24) + B(x^2 - 2x - 48) + C(x^2 + 3x - 18) \]

Now, solve for \( A \), \( B \), and \( C \). Once you have those values, integrate each term separately. Then, you can integrate the entire expression. Let me know if you need further assistance with solving for \( A \), \( B \), and \( C \). Ayan lang alam ko.