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Đáp án:

$\begin{array}{l}
a)\int {\dfrac{{2{x^2} - 3x - 5}}{{x - 2}}dx} \\
 = \int {\dfrac{{2x\left( {x - 2} \right) + x - 2 - 3}}{{x - 2}}dx} \\
 = \int {2x + 1 - \dfrac{3}{{x - 2}}dx} \\
 = {x^2} + x - 3\ln \left| {x - 2} \right| + C\\
b)\int {\dfrac{{2x + 1}}{{x - 3}}dx} \\
 = \int {\dfrac{{2\left( {x - 3} \right) + 7}}{{x - 3}}dx} \\
 = \int {2 + \dfrac{7}{{x - 3}}dx} \\
 = 2x + 7\ln \left| {x - 3} \right| + C\\
c)\int {\dfrac{{4x - 5}}{{2x - 1}}dx} \\
 = \int {\dfrac{{2\left( {2x - 1} \right) - 3}}{{2x - 1}}dx} \\
 = \int {2 - \dfrac{3}{{2x - 1}}dx} \\
 = 2x - \dfrac{3}{2}\ln \left| {2x - 1} \right| + C\\
d)\int {\dfrac{{dx}}{{\left( {x + 1} \right)\left( {2x - 3} \right)}}} \\
 = \int {\dfrac{a}{{2x - 3}} - \dfrac{b}{{x + 1}}dx} \\
 = \int {\dfrac{2}{{5\left( {2x - 3} \right)}} - \dfrac{1}{{5\left( {x + 1} \right)}}dx} \\
 = \dfrac{1}{5}\ln \left| {2x - 3} \right| - \dfrac{1}{5}\ln \left| {x + 1} \right| + C\\
e)\int {\dfrac{{3x + 1}}{{{x^2} - 4}}dx} \\
 = \int {\dfrac{{3x}}{{{x^2} - 4}} + \dfrac{1}{{{x^2} - 4}}dx} \\
 = \int {\dfrac{3}{2}.\dfrac{{2x}}{{{x^2} - 4}}dx}  + \int {\dfrac{1}{{{x^2} - 4}}dx} \\
 = \dfrac{3}{2}.\ln \left| {{x^2} - 4} \right| + \int {\dfrac{1}{{4\left( {{{\sin }^2}x - 1} \right)}}.2\cos xdx} \\
\left( {x = 2\sin x \Rightarrow dx = 2\cos xdx} \right)\\
 = \dfrac{3}{2}\ln \left| {{x^2} - 4} \right| - \int {\dfrac{1}{2}.\dfrac{1}{{{{\cos }^2}x}}.\cos xdx} \\
 = \dfrac{3}{2}\ln \left| {{x^2} - 4} \right| - \dfrac{1}{2}\int {.\dfrac{1}{{\cos x}}dx} \\
 = \dfrac{3}{2}\ln \left| {{x^2} - 4} \right| - \dfrac{1}{2}.{\cot ^2}x + C\\
f)\int {\dfrac{1}{{{x^2} + 3x - 4}}dx} \\
 = \int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 4} \right)}}dx} \\
 = \int {\dfrac{1}{5}.\dfrac{{x + 4 - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 4} \right)}}dx} \\
 = \dfrac{1}{5}.\int {\dfrac{1}{{x - 1}} - \dfrac{1}{{x + 4}}dx} \\
 = \dfrac{1}{5}.\left( {\ln \left| {x - 1} \right| - \ln \left| {x + 4} \right|} \right) + C
\end{array}$