Answer: (b) \(\frac32\ln(4x^2-9)+c\); (c) \(a=\sqrt{11}\).

(a) Use the common denominator

\((2x-3)(2x+3)=4x^2-9.\)

Then

\(\frac{3}{2x-3}+\frac{3}{2x+3} =\frac{3(2x+3)+3(2x-3)}{4x^2-9}.\)

The numerator simplifies to

\(6x+9+6x-9=12x.\)

So

\(\frac{3}{2x-3}+\frac{3}{2x+3} =\frac{12x}{4x^2-9}.\)

(b) Hence

\(\int\frac{12x}{4x^2-9}\,dx =\int\left(\frac{3}{2x-3}+\frac{3}{2x+3}\right)\,dx.\)

This gives

\(\frac32\ln(2x-3)+\frac32\ln(2x+3)+c.\)

Combine the logarithms:

\(\frac32\ln\left((2x-3)(2x+3)\right)+c.\)

Therefore

\(\int\frac{12x}{4x^2-9}\,dx =\frac32\ln(4x^2-9)+c.\)

(c) Using the result from part (b),

\(\int_2^a\frac{12x}{4x^2-9}\,dx =\frac32\ln(4a^2-9)-\frac32\ln(4(2)^2-9).\)

Since \(4(2)^2-9=7\), this is

\(\frac32\ln(4a^2-9)-\frac32\ln7.\)

So

\(\frac32\ln\left(\frac{4a^2-9}{7}\right)=\ln(5\sqrt5).\)

Now \(5\sqrt5=5^{3/2}\), so

\(\ln(5\sqrt5)=\frac32\ln5.\)

Hence

\(\ln\left(\frac{4a^2-9}{7}\right)=\ln5.\)

Therefore

\(\frac{4a^2-9}{7}=5.\)

So

\(4a^2-9=35,\)

and

\(a^2=11.\)

Since \(a\gt2\),

\(a=\sqrt{11}.\)