Đáp án:
$\begin{array}{l}
k)\int {x.\ln xdx} \\
\left\{ \begin{array}{l}
u = \ln x \Rightarrow du = \frac{{dx}}{x}\\
dv = xdx \Rightarrow v = \frac{{{x^2}}}{2}
\end{array} \right.\\
\Rightarrow I = u.v - \int {v.du} \\
= {\mathop{\rm lnx}\nolimits} .\frac{{{x^2}}}{2} - \int {\frac{1}{x}.\frac{{{x^2}}}{2}dx} \\
= \ln x.\frac{{{x^2}}}{2} - \frac{{{x^2}}}{4} + C\\
r)\int {x{{.2}^x}dx} \\
= \frac{{{2^x}}}{{\ln 2}}.x - \int {\frac{{{2^x}}}{{\ln 2}}dx} \\
= \frac{{{2^x}}}{{\ln 2}}.x - \frac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}} + C\\
i)\int {\ln xdx} \\
= x.\ln x - \int {x.\frac{1}{x}dx} \\
= x.\ln x - x + C
\end{array}$
