Integralen av $$$\frac{1}{a^{2} - u^{2}}$$$ med avseende på $$$u$$$

Bestäm $$$\int \frac{1}{a^{2} - u^{2}}\, du$$$.

Utför partialbråksuppdelning:

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

Integrera termvis:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2 a}$$$ och $$$f{\left(u \right)} = \frac{1}{a + u}$$$:

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

Låt $$$v=a + u$$$ vara.

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

Integralen blir

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

Integralen av $$$\frac{1}{v}$$$ är $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Kom ihåg att $$$v=a + u$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2 a}$$$ och $$$f{\left(u \right)} = \frac{1}{- a + u}$$$:

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

Låt $$$v=- a + u$$$ vara.

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

Integralen kan omskrivas som

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

Integralen av $$$\frac{1}{v}$$$ är $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Kom ihåg att $$$v=- a + u$$$:

$$\frac{\ln{\left(\left|{a + u}\right| \right)}}{2 a} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2 a} = \frac{\ln{\left(\left|{a + u}\right| \right)}}{2 a} - \frac{\ln{\left(\left|{{\color{red}{\left(- a + u\right)}}}\right| \right)}}{2 a}$$

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

$$$\int \frac{1}{a^{2} - u^{2}}\, du = \frac{- \ln\left(\left|{a - u}\right|\right) + \ln\left(\left|{a + u}\right|\right)}{2 a} + C$$$A