Integralen av $$$\frac{- b + x}{\sqrt{- a + x}}$$$ med avseende på $$$x$$$

Bestäm $$$\int \frac{- b + x}{\sqrt{- a + x}}\, dx$$$.

Skriv om täljaren till $$$- b + x=- b + x$$$ och dela upp bråket:

$${\color{red}{\int{\frac{- b + x}{\sqrt{- a + x}} d x}}} = {\color{red}{\int{\left(\sqrt{- a + x} + \frac{a - b}{\sqrt{- a + x}}\right)d x}}}$$

Integrera termvis:

$${\color{red}{\int{\left(\sqrt{- a + x} + \frac{a - b}{\sqrt{- a + x}}\right)d x}}} = {\color{red}{\left(\int{\frac{a - b}{\sqrt{- a + x}} d x} + \int{\sqrt{- a + x} d x}\right)}}$$

Låt $$$u=- a + x$$$ vara.

Då $$$du=\left(- a + x\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Integralen kan omskrivas som

$$\int{\frac{a - b}{\sqrt{- a + x}} d x} + {\color{red}{\int{\sqrt{- a + x} d x}}} = \int{\frac{a - b}{\sqrt{- a + x}} d x} + {\color{red}{\int{\sqrt{u} d u}}}$$

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

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

Kom ihåg att $$$u=- a + x$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=a - b$$$ och $$$f{\left(x \right)} = \frac{1}{\sqrt{- a + x}}$$$:

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

Låt $$$u=- a + x$$$ vara.

Då $$$du=\left(- a + x\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

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

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

Kom ihåg att $$$u=- a + x$$$:

$$\frac{2 \left(- a + x\right)^{\frac{3}{2}}}{3} + 2 \left(a - b\right) \sqrt{{\color{red}{u}}} = \frac{2 \left(- a + x\right)^{\frac{3}{2}}}{3} + 2 \left(a - b\right) \sqrt{{\color{red}{\left(- a + x\right)}}}$$

Alltså,

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

Förenkla:

$$\int{\frac{- b + x}{\sqrt{- a + x}} d x} = \frac{2 \sqrt{- a + x} \left(2 a - 3 b + x\right)}{3}$$

Lägg till integrationskonstanten:

$$\int{\frac{- b + x}{\sqrt{- a + x}} d x} = \frac{2 \sqrt{- a + x} \left(2 a - 3 b + x\right)}{3}+C$$

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