Intégrale de ln(x2)\ln\left(x^{2}\right)

La calculatrice trouvera l'intégrale/antidérivée de ln(x2)\ln\left(x^{2}\right), avec les étapes indiquées.

Calculatrice associée: Calculatrice intégrale

Solution

The input is rewritten: ln(x2)dx=2ln(x)dx\int{\ln{\left(x^{2} \right)} d x}=\int{2 \ln{\left(x \right)} d x}.

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

2ln(x)dx=(2ln(x)dx){\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 ln(x)dx\int{\ln{\left(x \right)} d x}, use integration by parts udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}.

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

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

Thus,

2ln(x)dx=2(ln(x)xx1xdx)=2(xln(x)1dx)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 cdx=cx\int c\, dx = c x with c=1c=1:

2xln(x)21dx=2xln(x)2x2 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,

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

Simplifier :

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

Ajouter la constante d'intégration :

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

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