Giải bài 3.14 trang 166 - SBT Đại số và Giải tích lớp 12
\(\int x\ln(x+1)dx\) bằng:
A. \((\dfrac{x^2}{2}-1)ln(x+1)+\dfrac{1}{4}(x-1)^2+C\)
B. \((\dfrac{x^2}{2}-1)\ln(x+1)-\dfrac{1}{2}(x-1)^2+C\)
C. \((\dfrac{x^2}{2}-\dfrac{1}{2})\ln(x+1)-\dfrac{1}{4}(x-1)^2+C\)
D. \((\dfrac{x^2}{2}+1)\ln(x+1)-\dfrac{1}{4}(x-1)^2+C\)
Đặt \(\left\{ \begin{aligned} & u=\ln \left( x+1 \right) \\ & dv=xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\frac{dx}{x+1} \\ & v=\frac{1}{2}{{x}^{2}} \\ \end{aligned} \right. \)
Ta có:
\(\begin{align} & \int{x\ln \left( x+1 \right)dx}=\frac{1}{2}{{x}^{2}}\ln \left( x+1 \right)-\frac{1}{2}\int{\frac{{{x}^{2}}}{x+1}dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}{{x}^{2}}\ln \left( x+1 \right)-\frac{1}{2}\int{\left( x-1+\frac{1}{x+1} \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}{{x}^{2}}\ln \left( x+1 \right)-\frac{1}{2}\left( \frac{{{x}^{2}}}{2}-x+\ln \left| x+1 \right| \right)+C \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{{{x}^{2}}}{2}-\frac{1}{2} \right)\ln \left( x+1 \right)-\frac{{{x}^{2}}}{4}+\frac{x}{2}+C \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{{{x}^{2}}}{2}-\frac{1}{2} \right)\ln \left( x+1 \right)-\frac{1}{4}{{\left( x-1 \right)}^{2}}+C \\ \end{align} \)
Chọn đáp án C.