Đáp án:
$\begin{array}{l}
a)f\left( x \right) = {\left( {1 - 2x} \right)^5}\\
\Rightarrow \int {f\left( x \right)dx} \\
= \int {{{\left( {1 - 2x} \right)}^5}dx} \\
= - \dfrac{1}{2}\int {{{\left( {1 - 2x} \right)}^5}.\left( { - 2} \right)dx} \\
= - \dfrac{1}{2}\int {{{\left( {1 - 2x} \right)}^5}d\left( {1 - 2x} \right)} \\
= - \dfrac{1}{2}.\dfrac{{{{\left( {1 - 2x} \right)}^6}}}{6} + C\\
= \dfrac{{ - {{\left( {2x - 1} \right)}^6}}}{{12}} + C\\
b)\int {f\left( x \right)dx} = \int {\dfrac{1}{{4{x^2} + 4x + 1}}dx} \\
= \int {\dfrac{1}{{{{\left( {2x + 1} \right)}^2}}}dx} \\
= \dfrac{1}{2}\int {{{\left( {2x + 1} \right)}^{ - 2}}.2dx} \\
= \dfrac{1}{2}\int {{{\left( {2x + 1} \right)}^{ - 2}}d\left( {2x + 1} \right)} \\
= \dfrac{1}{2}.\left( { - 1} \right).{\left( {2x + 1} \right)^{ - 1}}d\left( {2x + 1} \right)\\
= - \dfrac{1}{{2\left( {2x + 1} \right)}} + C\\
c)\int {f\left( x \right)d{\rm{x}}} \\
= \int {\dfrac{3}{{2 - 3x}}dx} \\
= - \int {\dfrac{1}{{2 - 3x}}\left( { - 3dx} \right)} \\
= - \int {\dfrac{1}{{2 - 3x}}d\left( {2 - 3x} \right)} \\
= - \ln \left| {2 - 3x} \right| + C\\
d)\int {f\left( x \right)dx} \\
= \int {{e^{ - x}}dx} \\
= - \int {{e^{ - x}}\left( { - dx} \right)} \\
= - {e^{ - x}} + C\\
e)\int {f\left( x \right)dx} \\
= \int {{e^{2x - 3}}dx} \\
= \dfrac{1}{2}.\int {{e^{2x - 3}}2dx} \\
= \dfrac{1}{2}\int {{e^{2x - 3}}.d\left( {2x - 3} \right)} \\
= \dfrac{1}{2}{e^{2x - 3}} + C\\
f)\int {f\left( x \right)dx} \\
= \int {\left( {1 - x} \right){e^{1 + 2x - {x^2}}}dx} \\
= \dfrac{1}{2}.\int {{e^{1 + 2x - {x^2}}}.\left( {2 - 2x} \right)dx} \\
= \dfrac{1}{2}\int {{e^{1 + 2x - {x^2}}}d\left( {1 + 2x - {x^2}} \right)} \\
= \dfrac{1}{2}{e^{\left( {1 + 2x - {x^2}} \right)}} + C
\end{array}$
