Integralen av $$$\operatorname{atan}{\left(2 x \right)}$$$

Bestäm $$$\int \operatorname{atan}{\left(2 x \right)}\, dx$$$.

Låt $$$u=2 x$$$ vara.

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

Integralen blir

$${\color{red}{\int{\operatorname{atan}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{2} d u}}}$$

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

$${\color{red}{\int{\frac{\operatorname{atan}{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\operatorname{atan}{\left(u \right)} d u}}{2}\right)}}$$

För integralen $$$\int{\operatorname{atan}{\left(u \right)} d u}$$$, använd partiell integration $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Låt $$$\operatorname{g}=\operatorname{atan}{\left(u \right)}$$$ och $$$\operatorname{dv}=du$$$.

Då gäller $$$\operatorname{dg}=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du=\frac{du}{u^{2} + 1}$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d u}=u$$$ (stegen kan ses »).

Integralen blir

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

Låt $$$v=u^{2} + 1$$$ vara.

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

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(v \right)} = \frac{1}{v}$$$:

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

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

$$\frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Kom ihåg att $$$v=u^{2} + 1$$$:

$$\frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{4}$$

Kom ihåg att $$$u=2 x$$$:

$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{4} + \frac{{\color{red}{u}} \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} = - \frac{\ln{\left(1 + {\color{red}{\left(2 x\right)}}^{2} \right)}}{4} + \frac{{\color{red}{\left(2 x\right)}} \operatorname{atan}{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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