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Using partial fractions formula found in MF15,

Let $\frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} = \frac{A}{2x-5} + \frac{Bx+C}{x^{2}+9}$

$9x^{2}+x-13 = A(x^{2}+9)+(Bx+C)(2x-5)$

$9 = A + 2B \longrightarrow 1$

$1 = 2C - 5B \longrightarrow 2$

$-13 = 9A - 5C \longrightarrow 3$

Using GC, $A = 3, B = 3, C = 8$

$\int_{0}^{2} \frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} dx$

$= \int_{0}^{2} \frac{3}{2x-5} + \frac{3x+8}{x^{2}+9} dx$

$= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3x}{x^{2}+9} + \frac{8}{x^{2}+9}dx$

$= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3}{2} \frac{2x}{x^{2}+9} + 8 \frac{1}{x^{2}+9}dx$

$= [\frac{3}{2} \mathrm{ln}|2x-5|+ \frac{3}{2} \mathrm{ln}(x^{2}+9) + \frac{8}{3} tan^{-1} \frac{8}{3}]\biggl|_0^2$

$= \frac{3}{2} \mathrm{ln} 13 + \frac{8}{3} tan^{-1} \frac{2}{3} - (\frac{3}{2} \mathrm{ln}5 + \frac{3}{2} \mathrm{ln}9]$

$= \frac{3}{2} \mathrm{ln} \frac{13}{45} + \frac{8}{3} tan^{-1} \frac{2}{3}$

$a = \frac{3}{2}, b = \frac{13}{45}, c = \frac{8}{3}, d = \frac{2}{3}$