Đáp án:
$\begin{array}{l}
1)I = \int\limits_1^2 {\dfrac{{5{x^3} - 4}}{{{x^3}}}dx} \\
= \int\limits_1^2 {5 - \dfrac{4}{{{x^3}}}dx} \\
= \left( {5x + \dfrac{4}{2}.{x^{ - 2}}} \right)\Bigg\vert_1^2\\
= \left( {5x + \dfrac{2}{{{x^2}}}} \right)\Bigg\vert_1^2\\
= \left( {5.2 + \dfrac{2}{{{5^2}}} - 5 - 2} \right)\\
= 3 + \dfrac{2}{{25}}\\
= \dfrac{{77}}{{25}}\\
2)a)\int\limits_0^1 {\dfrac{{3{x^2}}}{{1 + {x^3}}}dx} \\
Đặt:{x^3} + 1 = a\\
\Rightarrow \left\{ \begin{array}{l}
da = 3{x^2}dx\\
x = 0 \Rightarrow a = 1\\
x = 1 \Rightarrow a = 2
\end{array} \right.\\
\Rightarrow I = \int\limits_1^2 {\dfrac{1}{a}da} = \left( {\ln a} \right)\Bigg\vert_1^2 = \ln 2\\
b)\int\limits_0^{\pi /2} {\dfrac{{\cos x}}{{1 + \sin x}}dx} \\
Đặt:1 + \sin x = a\\
\Rightarrow \left\{ \begin{array}{l}
da = \cos xdx\\
x = 0 \Rightarrow a = 1\\
x = \dfrac{\pi }{2} \Rightarrow a = 2
\end{array} \right.\\
\Rightarrow I = \int\limits_1^2 {\dfrac{{da}}{a}} = \left( {\ln a} \right)\Bigg\vert_1^2 = \ln 2\\
c)\int\limits_1^e {\dfrac{{{{\ln }^2}x}}{x}dx} \\
Đặt:\ln x = a \Rightarrow \left\{ \begin{array}{l}
da = \dfrac{1}{x}dx\\
x = 1 \Rightarrow a = 0\\
x = e \Rightarrow a = 1
\end{array} \right.\\
\Rightarrow I = \int\limits_0^1 {{a^2}da} = \left( {\dfrac{1}{3}{a^3}} \right)\Bigg\vert_0^1 = \dfrac{1}{3}\\
4)\int\limits_1^4 {\dfrac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} \\
Đặt:\sqrt x = a\\
\Rightarrow \left\{ \begin{array}{l}
da = \dfrac{1}{{2\sqrt x }}dx \Rightarrow \dfrac{1}{{\sqrt x }}dx = 2da\\
x = 1 \Rightarrow a = 1\\
x = 4 \Rightarrow a = 2
\end{array} \right.\\
\Rightarrow I = \int\limits_1^2 {{e^a}.2da} \\
= \left( {2{e^a}} \right)\Bigg\vert_1^2 = 2{e^2} - 2e
\end{array}$
