Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
t = \ln x \Rightarrow dt = \left( {\ln x} \right)'.dx = \frac{1}{x}dx\\
\int {\frac{{3{{\ln }^2}x + 6\ln x + 3}}{x}dx} = \int {\left( {3{{\ln }^2}x + 6\ln x + 3} \right)\frac{{dx}}{x}} \\
= \int {\left( {3{t^2} + 6t + 3} \right)dt} \\
= {t^3} + 3{t^2} + 3t + C\\
= {\ln ^3}x + 3{\ln ^2}x + 3\ln x + C\\
3,\\
\int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}x\cos xdx} = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}x.\left( {\cos xdx} \right)} = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}xd\left( {\sin x} \right)} \\
= \mathop {\left. {\frac{{{{\sin }^5}x}}{5}} \right|}\nolimits_0^{\frac{\pi }{2}} = \frac{1}{5}
\end{array}\)
