Answer: \(a=3\).

Integrate term by term:

\(\int\left(\frac2{2x+3}+\frac3{3x-1}-\frac1x\right)\,dx =\ln(2x+3)+\ln(3x-1)-\ln x.\)

So

\(\left[\ln(2x+3)+\ln(3x-1)-\ln x\right]_1^a=\ln2.4.\)

Substitute the limits:

\(\ln(2a+3)+\ln(3a-1)-\ln a-(\ln5+\ln2)=\ln2.4.\)

Using logarithm laws,

\(\ln\left(\frac{(2a+3)(3a-1)}{10a}\right)=\ln2.4.\)

Therefore

\(\frac{(2a+3)(3a-1)}{10a}=2.4.\)

Since \(2.4=\frac{12}{5}\),

\((2a+3)(3a-1)=24a.\)

Expand:

\(6a^2+7a-3=24a.\)

So

\(6a^2-17a-3=0.\)

Use the quadratic formula:

\(a=\frac{17\pm\sqrt{(-17)^2-4(6)(-3)}}{12} =\frac{17\pm\sqrt{361}}{12}.\)

Thus

\(a=\frac{17\pm19}{12}.\)

So

\(a=3\quad\text{or}\quad a=-\frac16.\)

Given that \(a\gt 1\),

\(a=3.\)