Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\int {\frac{{(2x - 3) + (4 - x) + (4 - x)}}{{(2x - 3)(4 - x)}}dx = \int {(\frac{1}{{4 - x}} + } } \frac{1}{{2x - 3}} + \frac{1}{{4 - x}})dx\\
= \int {\left( {\frac{2}{{4 - x}} + \frac{1}{{2x - 3}}} \right)dx = \int {\left( {\frac{2}{{4 - x}}} \right)dx} } + \int {\left( {\frac{1}{{2x - 3}}} \right)dx} \\
\end{array}\)
Đặt 4-x=t⇒-dt=dx
2x+3=u⇒dx=du/3
\(\begin{array}{l}
\to \int {\left( {\frac{2}{{4 - x}}} \right)dx} + \int {\left( {\frac{1}{{2x - 3}}} \right)dx} = - 2\int {\frac{1}{t}dt + \frac{1}{3}\int {\frac{1}{u}du} } \\
= - 2\ln \left| t \right| + \frac{1}{3}\ln \left| u \right| + C = - 2\ln \left| {4 - x} \right| + \frac{1}{3}\ln \left| {2x + 3} \right| + C
\end{array}\)
