Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
I = \int {\cos 2xdx} \\
t = 2x \Rightarrow dt = \left( {2x} \right)'.dx = 2dx\\
\Rightarrow I = \int {\cos t.\frac{{dt}}{2}} = \frac{1}{2}\int {\cos t.dt} = \frac{1}{2}\sin t + C = \frac{1}{2}\sin 2x + C\\
*)\\
I = \int {\left( {\frac{1}{x} - 2x} \right)dx} = \int {\frac{{dx}}{x}} - \int {2xdx} = \ln \left| x \right| - {x^2} + C\\
*)\\
I = \int {\frac{1}{{{x^2} - 4x + 4}}dx} = \int {\frac{1}{{{{\left( {x - 2} \right)}^2}}}dx} \\
t = x - 2 \Rightarrow dt = \left( {x - 2} \right)'.dx = 1.dx = dx\\
\Rightarrow I = \int {\frac{1}{{{t^2}}}dt} = - \frac{1}{t} + C = - \frac{1}{{x - 2}} + C = \frac{1}{{2 - x}} + C\\
*)\\
I = \int\limits_1^4 {\frac{1}{{2\sqrt x }}dx} = \mathop {\left. {\sqrt x } \right|}\nolimits_1^4 = \sqrt 4 - \sqrt 1 = 1
\end{array}\)
