Intégrale de $\ln\left(x^{2}\right)$

Trouvez $\int \ln\left(x^{2}\right)\, dx$.

The input is rewritten: $\int{\ln{\left(x^{2} \right)} d x}=\int{2 \ln{\left(x \right)} d x}$.

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=2$ and $f{\left(x \right)} = \ln{\left(x \right)}$:

$${\color{red}{\int{2 \ln{\left(x \right)} d x}}} = {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}$$

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 »).

The integral can be rewritten as

$$2 {\color{red}{\int{\ln{\left(x \right)} d x}}}=2 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=2 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

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

$$2 x \ln{\left(x \right)} - 2 {\color{red}{\int{1 d x}}} = 2 x \ln{\left(x \right)} - 2 {\color{red}{x}}$$

C'est pourquoi,

$$\int{2 \ln{\left(x \right)} d x} = 2 x \ln{\left(x \right)} - 2 x$$

Simplifier :

$$\int{2 \ln{\left(x \right)} d x} = 2 x \left(\ln{\left(x \right)} - 1\right)$$

Ajouter la constante d'intégration :

$$\int{2 \ln{\left(x \right)} d x} = 2 x \left(\ln{\left(x \right)} - 1\right)+C$$

Answer: $\int{2 \ln{\left(x \right)} d x}=2 x \left(\ln{\left(x \right)} - 1\right)+C$