Integralen av $$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$

Bestäm $$$\int \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\, dx$$$.

För integralen $$$\int{\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} d x}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Låt $$$\operatorname{u}=\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$ och $$$\operatorname{dv}=dx$$$.

Då gäller $$$\operatorname{du}=\left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right)^{\prime }dx=\frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right) \sqrt{x^{4} - 2 x^{2} + 1}} dx$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d x}=x$$$ (stegen kan ses »).

Integralen kan omskrivas som

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

Låt $$$u=x^{2} + 1$$$ vara.

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

Alltså,

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

Förenkla integranden:

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

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

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

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

$$x \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} + {\color{red}{\int{\frac{1}{u} d u}}} = x \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x^{2} + 1$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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