Đáp án: C
Giải thích các bước giải:
$\begin{array}{l}
\int {x\ln \left( {1 - x} \right)dx} \\
\left\{ \begin{array}{l}
u = \ln \left( {1 - x} \right) \Rightarrow du = \frac{{ - 1}}{{1 - x}} = \frac{1}{{x - 1}}\\
dv = x \Rightarrow v = \frac{{{x^2}}}{2} - \frac{1}{2}
\end{array} \right.\\
\Rightarrow I = u.v - \int {v.du} \\
= \left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right).\ln \left( {1 - x} \right) - \int {\left( {\frac{{{x^2} - 1}}{2}} \right).\frac{1}{{x - 1}}du} \\
= \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\ln \left( {1 - x} \right) - \int {\frac{{x + 1}}{2}dx} \\
= \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\ln \left( {1 - x} \right) - \left( {\frac{{{x^2}}}{4} + \frac{1}{2}x} \right)\\
= \frac{{{x^2}}}{2}\ln \left( {1 - x} \right) - \frac{1}{2}\ln \left( {1 - x} \right) - \frac{1}{4}{\left( {x + 1} \right)^2} + C\\
\Rightarrow \left\{ \begin{array}{l}
m = 2\\
n = 2\\
k = 4
\end{array} \right. \Rightarrow m - n + k = 4
\end{array}$
