Phương pháp giải:

\(\frac{1}{{{x}^{2}}-x-12}=\frac{1}{\left( x-4 \right)\left( x+3 \right)}=\frac{A}{x-4}+\frac{B}{x+3}\)

Lời giải chi tiết:

Ta có : \(\frac{1}{{{x}^{2}}-x-12}=\frac{1}{\left( x-4 \right)\left( x+3 \right)}=\frac{1}{7}\left( \frac{1}{x-4}-\frac{1}{x+3} \right)\)

\(\Rightarrow I=\frac{1}{7}\int\limits_{0}^{1}{\left( \frac{1}{x-4}-\frac{1}{x+3} \right)dx}=\left. \frac{1}{7}\ln \left| \frac{x-4}{x+3} \right| \right|_{0}^{1}=\frac{1}{7}\left( \ln \frac{3}{4}-\ln \frac{4}{3} \right)=\frac{1}{7}\ln \frac{9}{16}\)

Chọn D.