Đáp án:
$\begin{array}{l}
\sqrt {{x^2} + 1} = a\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = {a^2} - 1\\
xdx = ada\\
x = 1 \Rightarrow a = \sqrt 2 \\
x = e \Rightarrow a = \sqrt {{e^2} + 1}
\end{array} \right.\\
\Rightarrow \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} - 1}}dx} \\
= \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} - 1}}.xdx} \\
= \int {\frac{{{a^2} - 1}}{{a - 1}}ada} \\
= \int {\left( {a + 1} \right)ada} \left( {do:a \ne 1} \right)\\
= \frac{1}{3}.{a^3} + \frac{1}{2}{a^2} + c\\
\Rightarrow \int\limits_1^e {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} - 1}}dx} \\
= \left( {\frac{1}{3}{a^3} + \frac{1}{2}{a^2}} \right)_{\sqrt 2 }^{\sqrt {{e^2} + 1} }\\
= \frac{1}{3}.\left( {{e^2} + 1} \right)\sqrt {{e^2} + 1} - \frac{{2\sqrt 2 }}{3} + \frac{1}{2}\left( {{e^2} + 1} \right) - 1
\end{array}$
