Đáp án:
$\begin{array}{l}
{I_4} = \int\limits_0^4 {x\sqrt {9 + {x^2}} dx} \\
Đặt:{x^2} + 9 = a\\
\Rightarrow 2xdx = da\\
\Rightarrow \left\{ \begin{array}{l}
xdx = \dfrac{{da}}{2}\\
x = 0 \Rightarrow a = 9\\
x = 4 \Rightarrow a = 25
\end{array} \right.\\
\Rightarrow {I_4} = \int\limits_9^{25} {\sqrt a .\dfrac{{da}}{2}} \\
= \left( {\dfrac{1}{2}.\dfrac{2}{3}.a\sqrt a } \right)\left| {_9^{25}} \right.\\
= \left( {\dfrac{1}{3}.25\sqrt {25} - \dfrac{1}{3}9\sqrt 9 } \right)\\
= \dfrac{{98}}{3}\\
{I_5} = \int\limits_0^{\ln 2} {{{\left( {{e^x} + 1} \right)}^3}{e^x}dx} \\
Đặt:{e^x} + 1 = a\\
\Rightarrow \left( {{e^x} + 1} \right)' = a'\\
\Rightarrow \left\{ \begin{array}{l}
{e^x}dx = da\\
x = 0 \Rightarrow a = 2\\
x = \ln 2 \Rightarrow a = 3
\end{array} \right.\\
\Rightarrow {I_5} = \int\limits_2^3 {{a^3}da} \\
= \left( {\dfrac{1}{4}{a^4}} \right)\left| {_2^3} \right.\\
= \dfrac{1}{4}\left( {{3^4} - {2^4}} \right)\\
= \dfrac{{65}}{4}
\end{array}$
