Integralen av $$$x \ln\left(x\right) - x$$$

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

Integrera termvis:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

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

För integralen $$$\int{x \ln{\left(x \right)} d x}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

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

Då gäller $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{x d x}=\frac{x^{2}}{2}$$$ (stegen kan ses »).

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(x \right)} = x$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

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

Alltså,

$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \ln{\left(x \right)}}{2} - \frac{3 x^{2}}{4}$$

Förenkla:

$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \left(2 \ln{\left(x \right)} - 3\right)}{4}$$

Lägg till integrationskonstanten:

$$\int{\left(x \ln{\left(x \right)} - x\right)d x} = \frac{x^{2} \left(2 \ln{\left(x \right)} - 3\right)}{4}+C$$

$$$\int \left(x \ln\left(x\right) - x\right)\, dx = \frac{x^{2} \left(2 \ln\left(x\right) - 3\right)}{4} + C$$$A