Integral of $\ln\left(x\right)$

Find $\int \ln\left(x\right)\, dx$.

For the integral $\int{\ln{\left(x \right)} d x}$, use integration by parts $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$.

Let $\operatorname{u}=\ln{\left(x \right)}$ and $\operatorname{dv}=dx$.

Then $\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$ (steps can be seen ») and $\operatorname{v}=\int{1 d x}=x$ (steps can be seen »).

Thus,

$${\color{red}{\int{\ln{\left(x \right)} d x}}}={\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}={\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

Apply the constant rule $\int c\, dx = c x$ with $c=1$:

$$x \ln{\left(x \right)} - {\color{red}{\int{1 d x}}} = x \ln{\left(x \right)} - {\color{red}{x}}$$

Therefore,

$$\int{\ln{\left(x \right)} d x} = x \ln{\left(x \right)} - x$$

Simplify:

$$\int{\ln{\left(x \right)} d x} = x \left(\ln{\left(x \right)} - 1\right)$$

Add the constant of integration:

$$\int{\ln{\left(x \right)} d x} = x \left(\ln{\left(x \right)} - 1\right)+C$$

$\int \ln\left(x\right)\, dx = x \left(\ln\left(x\right) - 1\right) + C$A