Hàm số f(x)=x(1-x)^10 có nguyên hàm là

Đáp án:

\[\int {f\left( x \right)dx}  =  - \frac{{{{\left( {1 - x} \right)}^{11}}\left( {11x + 1} \right)}}{{132}} + C\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
f\left( x \right) = \int {x{{\left( {1 - x} \right)}^{10}}} dx\\
\left\{ \begin{array}{l}
u = x\\
v' = {\left( {1 - x} \right)^{10}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = \frac{{ - 1}}{{11}}.{\left( {1 - x} \right)^{11}}
\end{array} \right.\\
 \Rightarrow f\left( x \right) = x.\left( { - \frac{1}{{11}}.{{\left( {1 - x} \right)}^{11}}} \right) - \int {\left( {\frac{{ - 1}}{{11}}} \right).{{\left( {1 - x} \right)}^{11}}dx} \\
 = \frac{{ - x{{\left( {1 - x} \right)}^{11}}}}{{11}} + \frac{1}{{11}}.\int {{{\left( {1 - x} \right)}^{11}}dx} \\
 = \frac{{ - x{{\left( {1 - x} \right)}^{11}}}}{{11}} + \frac{1}{{11}}.\left( { - \frac{1}{{12}}{{\left( {1 - x} \right)}^{12}}} \right) + C\\
 =  - \frac{{{{\left( {1 - x} \right)}^{11}}}}{{11}}\left( {x + \frac{1}{{12}}\left( {1 - x} \right)} \right) + C\\
 =  - \frac{{{{\left( {1 - x} \right)}^{11}}}}{{11}}\left( {\frac{{11}}{{12}}x + \frac{1}{{12}}} \right) + C\\
 =  - \frac{{{{\left( {1 - x} \right)}^{11}}\left( {11x + 1} \right)}}{{132}} + C
\end{array}\)