Integralen av $$$4 x + \frac{3}{x^{2}}$$$

Bestäm $$$\int \left(4 x + \frac{3}{x^{2}}\right)\, dx$$$.

Integrera termvis:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=3$$$ och $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = x$$$:

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

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

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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