$\int\limits^1_0 ln \, dx$

$\displaystyle\int\limits^1_0 \ln x \, dx=\displaystyle\lim_{a \to 0}\displaystyle\int\limits^1_a \ln x \, dx \\ u=\ln x \Rightarrow du=\dfrac{dx}{x}\\ dv=dx \Rightarrow v=x\\ I=\displaystyle\lim_{a \to 0}\left(x\ln x \Bigg\vert^1_a -\displaystyle\int\limits^1_a \, dx\right)\\ =\displaystyle\lim_{a \to 0}\left(-a\ln(a)-1+a\right)=-1$

Đáp án:

$\displaystyle\int\limits_0^1\ln xdx=-1$

Giải thích các bước giải:

$\quad \displaystyle\int\limits_0^1\ln xdx$

Đặt $\begin{cases}u = \ln x\\dv = dx\end{cases}\longrightarrow \begin{cases}du =\dfrac1xdx\\v = x\end{cases}$

Ta được:

$\quad \lim\limits_{t\to 0^+}x\ln x\Bigg|_t^1 - \displaystyle\int\limits_0^1 dx$

$=0 - x\Bigg|_0^1$

$= -1$