Đáp án:
\[I = \frac{{16467}}{{40}} + \ln 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_1^2 {\frac{{{{\left( {{x^2} + 1} \right)}^5}}}{x}dx} = \int\limits_1^2 {\frac{{{{\left( {{x^2} + 1} \right)}^5}}}{{2{x^2}}}.\left( {2xdx} \right)} \\
t = {x^2} + 1 \Rightarrow \left\{ \begin{array}{l}
{x^2} = t - 1\\
dt = \left( {{x^2} + 1} \right)'dx = 2xdx\\
x = 1 \Rightarrow t = 2\\
x = 2 \Rightarrow t = 5
\end{array} \right.\\
\Rightarrow I = \int\limits_2^5 {\frac{{{t^5}}}{{2\left( {t - 1} \right)}}dt} \\
= \frac{1}{2}.\int\limits_2^5 {\frac{{\left( {{t^5} - 1} \right) + 1}}{{t - 1}}dt} \\
= \frac{1}{2}.\int\limits_2^5 {\frac{{\left( {t - 1} \right)\left( {{t^4} + {t^3} + {t^2} + t + 1} \right) + 1}}{{\left( {t - 1} \right)}}dt} \\
= \frac{1}{2}.\int\limits_2^5 {\left( {{t^4} + {t^3} + {t^2} + t + 1 + \frac{1}{{t - 1}}} \right)dt} \\
= \frac{1}{2}.\mathop {\left. {\left( {\frac{1}{5}{t^5} + \frac{1}{4}{t^4} + \frac{1}{3}{t^3} + \frac{1}{2}{t^2} + t + \ln \left| {t - 1} \right|} \right)} \right|}\nolimits_2^5 \\
= \frac{1}{2}.\left( {\frac{{16467}}{{20}} + \ln 4 - \ln 1} \right)\\
= \frac{{16467}}{{40}} + \ln 2
\end{array}\)
