The partial fraction decomposition for the rational expression (x-4)/(3(7x^2+3)) can be written in the form:
(x-4)/(3(7x^2+3)) = A/(7x+√3) + B/(7x-√3)
where A and B are constants to be determined.
To find A and B, we can multiply both sides of the equation by the denominator 3(7x^2+3) and then substitute values of x that make one of the denominators zero. This gives us two equations:
A(7x-√3) + B(7x+√3) = x-4
A(7x+√3) + B(7x-√3) = x-4
Solving these equations simultaneously, we get:
A = (1 + 4√3)/(42√3)
B = (1 - 4√3)/(42√3)
Now we can integrate the partial fraction decomposition:
∫(x-4)/(3(7x^2+3)) dx = (1/(42√3)) ∫(1/(7x+√3) + 1/(7x-√3)) dx
Using the substitution u = 7x + √3 and du/dx = 7, we get:
(1/(42√3)) ∫(1/(7x+√3) + 1/(7x-√3)) dx = (1/(42√3)) ln|7x+√3| - (1/(42√3)) ln|7x-√3| + C
where C is the constant of integration.