Đáp án:
c) $\dfrac{{{e^{{x^2}}}}}{2}\left( {{x^2} - 1} \right) + C$
d) $2{e^{\sqrt x }}\left( {\sqrt x - 1} \right) + C$
Giải thích các bước giải:
c) \(I = \int {{x^3}{e^{{x^2}}}dx} \)
Đặt \({x^2} = t \Rightarrow 2xdx = dt \Rightarrow xdx = \dfrac{1}{2}dt\)
\( \Rightarrow I = \int {{x^2}{e^{{x^2}}}.xdx} = \int {t.{e^t}.\dfrac{1}{2}dt} = \dfrac{1}{2}\int {t{e^t}dt} \)
Đặt \(\left\{ \begin{array}{l}t = u\\{e^t}dt = dv\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right.\)
\( \Rightarrow I = \dfrac{1}{2}\int {t{e^t}dt} = \dfrac{1}{2}t{e^t} - \dfrac{1}{2}\int {{e^t}dt} = \dfrac{1}{2}t{e^t} - \dfrac{1}{2}{e^t} + C\)
\( = \dfrac{{{e^t}}}{2}\left( {t - 1} \right) + C = \dfrac{{{e^{{x^2}}}}}{2}\left( {{x^2} - 1} \right) + C\)
d) \(I = \int {{e^{\sqrt x }}dx} \)
Đặt \(\sqrt x = t \Rightarrow x = {t^2} \Rightarrow dx = 2tdt\)
\( \Rightarrow I = \int {{e^t}.2tdt} = 2\int {t{e^t}dt} \)
Đặt \(\left\{ \begin{array}{l}t = u\\{e^t}dt = dv\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right.\)
\( \Rightarrow I = 2\int {t{e^t}dt} = 2t{e^t} - 2\int {{e^t}dt} = 2t{e^t} - 2{e^t} + C\)
\( = 2{e^t}\left( {t - 1} \right) + C = 2{e^{\sqrt x }}\left( {\sqrt x - 1} \right) + C\)
