Funktion $$$\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3}$$$ integraali

Määritä $$$\int \frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3}\, dx$$$.

Olkoon $$$u=x - 3$$$.

Tällöin $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = du$$$.

Näin ollen,

$${\color{red}{\int{\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3} d x}}} = {\color{red}{\int{\frac{\left(u - 1\right) \left(u + 1\right) \left(u + 2\right)}{u} d u}}}$$

Expand the expression:

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

Integroi termi kerrallaan:

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

Sovella vakiosääntöä $$$\int c\, du = c u$$$ käyttäen $$$c=1$$$:

$$- \int{\frac{2}{u} d u} + \int{2 u d u} + \int{u^{2} d u} - {\color{red}{\int{1 d u}}} = - \int{\frac{2}{u} d u} + \int{2 u d u} + \int{u^{2} d u} - {\color{red}{u}}$$

Sovella potenssisääntöä $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=2$$$:

$$- u - \int{\frac{2}{u} d u} + \int{2 u d u} + {\color{red}{\int{u^{2} d u}}}=- u - \int{\frac{2}{u} d u} + \int{2 u d u} + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u - \int{\frac{2}{u} d u} + \int{2 u d u} + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=2$$$ ja $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\frac{u^{3}}{3} - u + \int{2 u d u} - {\color{red}{\int{\frac{2}{u} d u}}} = \frac{u^{3}}{3} - u + \int{2 u d u} - {\color{red}{\left(2 \int{\frac{1}{u} d u}\right)}}$$

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{u^{3}}{3} - u + \int{2 u d u} - 2 {\color{red}{\int{\frac{1}{u} d u}}} = \frac{u^{3}}{3} - u + \int{2 u d u} - 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=2$$$ ja $$$f{\left(u \right)} = u$$$:

$$\frac{u^{3}}{3} - u - 2 \ln{\left(\left|{u}\right| \right)} + {\color{red}{\int{2 u d u}}} = \frac{u^{3}}{3} - u - 2 \ln{\left(\left|{u}\right| \right)} + {\color{red}{\left(2 \int{u d u}\right)}}$$

Sovella potenssisääntöä $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=1$$$:

$$\frac{u^{3}}{3} - u - 2 \ln{\left(\left|{u}\right| \right)} + 2 {\color{red}{\int{u d u}}}=\frac{u^{3}}{3} - u - 2 \ln{\left(\left|{u}\right| \right)} + 2 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=\frac{u^{3}}{3} - u - 2 \ln{\left(\left|{u}\right| \right)} + 2 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Muista, että $$$u=x - 3$$$:

$$- 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - {\color{red}{u}} + {\color{red}{u}}^{2} + \frac{{\color{red}{u}}^{3}}{3} = - 2 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} - {\color{red}{\left(x - 3\right)}} + {\color{red}{\left(x - 3\right)}}^{2} + \frac{{\color{red}{\left(x - 3\right)}}^{3}}{3}$$

Näin ollen,

$$\int{\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3} d x} = - x + \frac{\left(x - 3\right)^{3}}{3} + \left(x - 3\right)^{2} - 2 \ln{\left(\left|{x - 3}\right| \right)} + 3$$

Sievennä:

$$\int{\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3} d x} = \frac{x^{3}}{3} - 2 x^{2} + 2 x - 2 \ln{\left(\left|{x - 3}\right| \right)} + 3$$

Lisää integraatiovakio (ja poista lausekkeesta vakio):

$$\int{\frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3} d x} = \frac{x^{3}}{3} - 2 x^{2} + 2 x - 2 \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \frac{\left(x - 4\right) \left(x - 2\right) \left(x - 1\right)}{x - 3}\, dx = \left(\frac{x^{3}}{3} - 2 x^{2} + 2 x - 2 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A