Đặt \(t = \sqrt {x + 1} \Rightarrow {t^2} = x + 1 \Rightarrow 2t{\rm{d}}t = {\rm{d}}x\).

Khi \(x = 3 \Rightarrow t = 2\); Khi \(x = 8 \Rightarrow t = 3\).

Khi đó  

\(\begin{array}{l} I = \int\limits_2^3 {\frac{1}{{{t^2} - 1 + \left( {{t^2} - 1} \right)t}}.2t} {\rm{d}}t = \int\limits_2^3 {\frac{{2t}}{{\left( {{t^2} - 1} \right)\left( {t + 1} \right)}}} {\rm{d}}t = \int\limits_2^3 {\frac{{2t}}{{\left( {t - 1} \right){{\left( {t + 1} \right)}^2}}}} {\rm{d}}t\\ = \int\limits_2^3 {\frac{{\left( {t + 1} \right) + \left( {t - 1} \right)}}{{\left( {t - 1} \right){{\left( {t + 1} \right)}^2}}}} {\rm{d}}t = \int\limits_2^3 {\left( {\frac{{\left( {t + 1} \right)}}{{\left( {t - 1} \right){{\left( {t + 1} \right)}^2}}} + \frac{{\left( {t - 1} \right)}}{{\left( {t - 1} \right){{\left( {t + 1} \right)}^2}}}} \right)} {\rm{d}}t\\ = \int\limits_2^3 {\left( {\frac{1}{{\left( {t - 1} \right)\left( {t + 1} \right)}} + \frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right)} {\rm{d}}t = \int\limits_2^3 {\left( {\frac{1}{2}.\frac{{\left( {t + 1} \right) - \left( {t - 1} \right)}}{{\left( {t - 1} \right)\left( {t + 1} \right)}} + \frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right)} {\rm{d}}t\\ = \int\limits_2^3 {\left[ {\frac{1}{2}\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right) + \frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right]} {\rm{d}}t = \left. {\left[ {\frac{1}{2}\left( {\ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} \right) - \frac{1}{{t + 1}}} \right]} \right|_2^3\\ = \left. {\left[ {\frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right| - \frac{1}{{t + 1}}} \right]} \right|_2^3 = \frac{1}{2}\ln \frac{1}{2} - \frac{1}{4} - \left( {\frac{1}{2}\ln \frac{1}{3} - \frac{1}{3}} \right)\\ = \frac{1}{2}\ln \frac{1}{2} - \frac{1}{2}\ln \frac{1}{3} - \frac{1}{4} + \frac{1}{3} = \frac{1}{2}\ln \frac{3}{2} + \frac{1}{{12}} \end{array}\)

⇒ a = 3, b = 2, c = 1, d = 12

Vậy abc - d = 3.2.1 - 12 =  - 6.