Bài tập 3.14 trang 166 SBT Toán 12
Bài tập 3.14 trang 166 SBT Toán 12
\(\mathop \smallint \nolimits x\ln (x + 1)dx\) bằng:
A. \(\left( {\frac{{{x^2}}}{2} - 1} \right)ln(x + 1) + \frac{1}{4}{(x - 1)^2} + C\)
B. \(\left( {\frac{{{x^2}}}{2} - 1} \right)\ln (x + 1) - \frac{1}{2}{(x - 1)^2} + C\)
C. \(\left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right)\ln (x + 1) - \frac{1}{4}{(x - 1)^2} + C\)
D. \(\left( {\frac{{{x^2}}}{2} + 1} \right)\ln (x + 1) - \frac{1}{4}{(x - 1)^2} + C\)
Đặt \(\left\{ \begin{array}{l}
u = ln\left( {x + 1} \right)\\
dv = xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{{x + 1}}\\
v = \frac{1}{2}{x^2}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\int {xln(x + 1)dx} = \frac{1}{2}{x^2}ln(x + 1) - \frac{1}{2}\int {\frac{{{x^2}}}{{x + 1}}} dx\\
= \frac{1}{2}{x^2}ln(x + 1) - \frac{1}{2}\int {\left( {x - 1 + \frac{1}{{x + 1}}dx} \right)} \\
= \frac{1}{2}{x^2}\ln \left( {x + 1} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right) + C\\
= \left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right)\ln \left( {x + 1} \right) - \frac{{{x^2}}}{4} + \frac{x}{2} + C\\
= \left( {\frac{{{x^2}}}{2} - \frac{1}{2}} \right)\ln \left( {x + 1} \right) - \frac{1}{4}{\left( {x - 1} \right)^2} + C
\end{array}\)
Chọn đáp án C.
-- Mod Toán 12
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