Integrate the ln

Relevant wiki: Integration by Parts - Easy

$\begin{aligned} I & = \int_0^1 \ln x \ dx & \small \color{#3D99F6}{\text{By integration by parts: }u = \ln x, \ v' = 1} \\ & = x \ln x \ \bigg|_0^1 - \int_0^1 \frac xx \ dx \\ & = 0 - \color{#3D99F6}{0} - x \ \bigg|_0^1 & \color{#3D99F6}{\text{Note that: }\lim_{x \to 0^+} x\ln x = 0 \text{, see Note.}} \\ & = - (1-0) \\ & = \boxed{-1} \end{aligned}$

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$\color{#3D99F6}{\text{Note:}}$

$\begin{aligned} L & = \lim_{x \to 0^+} x\ln x & \small \color{#3D99F6}{\text{Divide up and down by }x} \\ & = \lim_{x \to 0^+} \frac {\ln x}{\frac 1x} & \small \color{#3D99F6}{\text{It is a }\infty / \infty \text{ case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0^+} \frac {\frac 1x}{-\frac 1{x^2}} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t. }x.} \\ & = \lim_{x \to 0^+} -x \\ & = \color{#3D99F6}{0} \end{aligned}$

About L'Hôpital's rule .

For $\int_{0}^{1} \ln(x)\,dx$

$\begin{aligned} \int_{0}^{1} \ln(x)\,dx &= x\ln(x) - x+ \color{grey}{\ce{C}} \quad\quad\quad\quad\quad\quad{\text{Apply integration by parts to get the indefinite}} \\&= -1-0 \\&= -1 \space \square \end{aligned}$

Take note : integration by parts: $\int uv' = uv-\int u'v$

$\large \text{ADIOS!!!}$

Using Byparts... We will get $x(logx -1)$ in limits $0 - 1$ . Then Putting these values we get. -1