Tính tích phân $I=\displaystyle\displaystyle\int\limits_{0}^{1}\dfrac{x-3}{x+1}\mathrm{\,d}x$.
![]() | $I=2-5\ln2$ |
![]() | $I=1-4\ln2$ |
![]() | $I=\dfrac{7}{2}-5\ln3$ |
![]() | $I=4\ln3-1$ |
Chọn phương án B.
$\begin{aligned}I&=\displaystyle\int\limits_{0}^{1}\dfrac{x-3}{x+1}\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}\left(1-\dfrac{4}{x+1}\right)\mathrm{\,d}x\\ &=\left(x-4\ln|x+1|\right)\bigg|_0^1=1-4\ln2.\end{aligned}$