Bài tập 3.12 trang 165 SBT Toán 12
Bài tập 3.12 trang 165 SBT Toán 12
\(\mathop \smallint \nolimits x{e^{2x}}dx\) bằng
A. \(\mathop \smallint \nolimits \frac{{{e^{2x}}(x - 2)}}{2} + C\)
B. \(\mathop \smallint \nolimits \frac{{{e^{2x}} + 1}}{2} + C\)
C . \(\mathop \smallint \nolimits \frac{{{e^{2x}}(x - 1)}}{2} + C\)
D . \(\mathop \smallint \nolimits \frac{{{e^{2x}}(2x - 1)}}{4} + C\)
Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = {e^{2x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\mathop \smallint \nolimits x{e^{2x}}dx = \frac{1}{2}x{e^{2x}} - \frac{1}{2}\mathop \smallint \nolimits {e^{2x}}dx = \frac{1}{2}x{e^{2x}} - \frac{1}{4}{e^{2x}} + C\\
= \frac{{{e^{2x}}(2x - 1)}}{4} + C
\end{array}\)
Chọn đáp án D
-- Mod Toán 12
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