Find the value of 2∫1 (6x+1) / (6x2-7x+2) dx, expressing your answer in the form mln(2) + nln(3), where m and n are integers.

This fraction can’t be integrated easily, but if we split it using partial fractions, these will be easier to integrate. To do this, we need to factoise the denominator, as this will follow the method of partial fractions:
²∫₁ (6x+1) / (6x²-7x+2) dx = ²∫₁ (6x+1) / ((3x-2)(2x-1)) dx
We can then use partial fractions to split this fraction into two that can be integrated, by using the variables A and B to represent expressions that would multiply together to make 6x+1 when put above the parts of the factoised denominator.
²∫₁ (6x+1) / ((3x-2)(2x-1)) dx = ²∫₁ A/(3x-2) + B/(2x-1)dx
Following the method of partial fractions, cross-multiply the fractions:
²∫₁ A/(3x-2) + B/(2x-1)dx = ²∫₁ A(2x-1)/(3x-2) + B(3x-2)/(2x-1) dx
                    = ²∫₁ (2Ax – A + 3Bx – 2B)/((3x-2)(2x-1)) dx
Therefore, we can say 2Ax – A + 3Bx – 2B = 6x+1, and so 2Ax + 3Bx = 6x and -A -2B = 1. We can then solve these simultaneous equations by elimination or substiution and find that B = -8 and A = 15. Therefore:
²∫₁ (6x+1) / ((3x-2)(2x-1)) dx = ²∫₁ A/(3x-2) + B/(2x-1) dx
                       = ²∫₁ 15/(3x-2) + -8/(2x-1) dx
                       = ²∫₁ 15/(3x-2) dx - ²∫₁ 8/(2x-1) dx
Using standard integrals, the integral of a fraction where the numerator is the derivative of the denominator is ln|denominator|. The fractions above are almost like this, if we rewrite them as:
²∫₁ 15/(3x-2) dx - ²∫₁ 8/(2x-1) dx = (5)²∫₁ 3/(3x-2) dx - (4)²∫₁ 2/(2x-1) dx
                          = 5[ln|3x-2|]²₁ – 4[ln|2x-1|]²₁
                          = 5(ln(4) – ln(1)) – 4(ln(3) – ln(1))
                          = 5ln(4) – 4ln(3)
To get this into the form required for the question, we can use the law of logs: log(a^b) = b log(a):
5ln(4) – 4ln(3) = 5ln(2²) – 4ln(3)
            = 5(2)ln(2) – 4ln(3)
            = 10ln(2) - 4ln(3)