Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
13,\\
I = \int {\frac{{2x - 3}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}dx} \\
 = \frac{1}{3}.\int {\frac{{6x - 9}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}dx} \\
 = \frac{1}{3}.\int {\frac{{\left( {x + 1} \right) + 5\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}dx} \\
 = \frac{1}{3}.\int {\left[ {\frac{1}{{x - 2}} + \frac{5}{{x + 1}}} \right]dx} \\
 = \frac{1}{3}.\left[ {\ln \left| {x - 2} \right| + 5.\ln \left| {x + 1} \right|} \right] + C\\
 = \frac{5}{3}\ln \left| {x + 1} \right| + \frac{1}{3}\ln \left| {x - 2} \right| + C\\
 \Rightarrow a = \frac{5}{3};\,\,\,b = \frac{1}{3} \Rightarrow a + b = 2\\
14,\\
I = \int {\frac{{4x + 11}}{{{x^2} + 5x + 6}}dx} \\
 = \int {\frac{{4x + 11}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}dx} \\
 = \int {\frac{{\left( {x + 2} \right) + 3\left( {x + 3} \right)}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}dx} \\
 = \int {\left[ {\frac{1}{{x + 3}} + \frac{3}{{x + 2}}} \right]dx} \\
 = \ln \left| {x + 3} \right| + 3\ln \left| {x + 2} \right| + C\\
 \Rightarrow a = 3;\,\,\,\,b = 1\\
 \Rightarrow P = {a^2} + ab + {b^2} = 13\\
15,\\
I = \int {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx}  =  - \frac{1}{{f\left( x \right)}} + C
\end{array}\)