Integralen av $$$\frac{1}{x \sqrt{x^{2} + 4 x + 1}}$$$

Bestäm $$$\int \frac{1}{x \sqrt{x^{2} + 4 x + 1}}\, dx$$$.

Låt $$$u=\frac{1}{x}$$$ vara.

Då $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (stegen kan ses »), och vi har att $$$\frac{dx}{x^{2}} = - du$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=-1$$$ och $$$f{\left(u \right)} = \frac{1}{\sqrt{u^{2} + 4 u + 1}}$$$:

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

Kvadratkomplettera (stegen kan ses »): $$$ u ^{2} + 4 u + 1 = \left( u + 2\right)^{2} - 3$$$:

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

Låt $$$v=u + 2$$$ vara.

Då $$$dv=\left(u + 2\right)^{\prime }du = 1 du$$$ (stegen kan ses »), och vi har att $$$du = dv$$$.

Integralen blir

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

Låt $$$v=\sqrt{3} \cosh{\left(w \right)}$$$ vara.

Då $$$dv=\left(\sqrt{3} \cosh{\left(w \right)}\right)^{\prime }dw = \sqrt{3} \sinh{\left(w \right)} dw$$$ (stegen kan ses »).

Det följer också att $$$w=\operatorname{acosh}{\left(\frac{\sqrt{3} v}{3} \right)}$$$.

Integranden blir

$$$\frac{1}{\sqrt{ v ^{2} - 3}} = \frac{1}{\sqrt{3 \cosh^{2}{\left( w \right)} - 3}}$$$

Använd identiteten $$$\cosh^{2}{\left( w \right)} - 1 = \sinh^{2}{\left( w \right)}$$$:

$$$\frac{1}{\sqrt{3 \cosh^{2}{\left( w \right)} - 3}}=\frac{\sqrt{3}}{3 \sqrt{\cosh^{2}{\left( w \right)} - 1}}=\frac{\sqrt{3}}{3 \sqrt{\sinh^{2}{\left( w \right)}}}$$$

Om vi antar att $$$\sinh{\left( w \right)} \ge 0$$$, erhåller vi följande:

$$$\frac{\sqrt{3}}{3 \sqrt{\sinh^{2}{\left( w \right)}}} = \frac{\sqrt{3}}{3 \sinh{\left( w \right)}}$$$

Integralen blir

$$- {\color{red}{\int{\frac{1}{\sqrt{v^{2} - 3}} d v}}} = - {\color{red}{\int{1 d w}}}$$

Tillämpa konstantregeln $$$\int c\, dw = c w$$$ med $$$c=1$$$:

$$- {\color{red}{\int{1 d w}}} = - {\color{red}{w}}$$

Kom ihåg att $$$w=\operatorname{acosh}{\left(\frac{\sqrt{3} v}{3} \right)}$$$:

$$- {\color{red}{w}} = - {\color{red}{\operatorname{acosh}{\left(\frac{\sqrt{3} v}{3} \right)}}}$$

Kom ihåg att $$$v=u + 2$$$:

$$- \operatorname{acosh}{\left(\frac{\sqrt{3} {\color{red}{v}}}{3} \right)} = - \operatorname{acosh}{\left(\frac{\sqrt{3} {\color{red}{\left(u + 2\right)}}}{3} \right)}$$

Kom ihåg att $$$u=\frac{1}{x}$$$:

$$- \operatorname{acosh}{\left(\frac{\sqrt{3} \left(2 + {\color{red}{u}}\right)}{3} \right)} = - \operatorname{acosh}{\left(\frac{\sqrt{3} \left(2 + {\color{red}{\frac{1}{x}}}\right)}{3} \right)}$$

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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