Integrate partial fractions with logarithm and arctangent terms

Write the fraction as $A/(2x-5)+(Bx+C)/(x^2+9)$. Then $9x^2+x-13=A(x^2+9)+(Bx+C)(2x-5)$. At $x=5/2$, $A=3$; at $x=0$, $-13=27-5C$, so $C=8$; at $x=1$, $-3=30-3(B+8)$, so $B=3$. Thus the integrand is $3/(2x-5)+(3x+8)/(x^2+9)$.

An antiderivative is $\tfrac32\ln|2x-5|+\tfrac32\ln(x^2+9)+\tfrac83\tan^{-1}(x/3)$. Evaluation from $0$ to $2$ gives $\tfrac32(\ln1-\ln5+\ln13-\ln9)+\tfrac83\tan^{-1}(2/3)=\tfrac32\ln(13/45)+\tfrac83\tan^{-1}(2/3)$. Hence $a=3/2$, $b=13/45$, $c=8/3$, $d=2/3$.