Đáp án:
\(\begin{array}{l}
13)\,I = \dfrac{{11}}{3} - 4\ln 2\\
14)\,I = - 2{e^{ - 4}} + 2{e^{ - 2}}\\
15)\,I = 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
13)\,Dat\,\sqrt {x - 1} = t \Rightarrow {t^2} = x - 1 \Rightarrow \left\{ \begin{array}{l}
2tdt = dx\\
x = {t^2} + 1
\end{array} \right.\\
x = 1 \Rightarrow t = 0\\
x = 2 \Rightarrow t = 1\\
I = \int\limits_0^1 {\dfrac{{{t^2} + 1}}{{1 + t}}.2tdt} = \int\limits_0^1 {2t\left( {t - 1 + \dfrac{2}{{t + 1}}} \right)dt} \\
= \int\limits_0^1 {\left( {2{t^2} - 2t + \dfrac{{4t}}{{t + 1}}} \right)dt} = \int\limits_0^1 {\left( {2{t^2} - 2t + 4 - \dfrac{4}{{t + 1}}} \right)dt} \\
= \left. {\left( {\dfrac{2}{3}{t^3} - {t^2} + 4t - 4ln\left| {t + 1} \right|} \right)} \right|_0^1 = \dfrac{{11}}{3} - 4\ln 2\\
14)\,\int\limits_1^2 {4.{e^{ - 2x}}dx} = \left. {\left( { - \dfrac{1}{2}.4.{e^{ - 2x}}} \right)} \right|_1^2 = - 2{e^{ - 4}} + 2{e^{ - 2}}\\
15)\,\int\limits_{ - 1}^1 {\left( {{e^x} - {e^{ - x}}} \right)dx} = \left. {\left( {{e^x} + {e^{ - x}}} \right)} \right|_{ - 1}^1 = e + {e^{ - 1}} - \left( {{e^{ - 1}} + e} \right) = 0
\end{array}\)
