Funktion $$$2^{\sqrt{x}}$$$ integraali

Määritä $$$\int 2^{\sqrt{x}}\, dx$$$.

Kannan vaihto:

$${\color{red}{\int{2^{\sqrt{x}} d x}}} = {\color{red}{\int{e^{\sqrt{x} \ln{\left(2 \right)}} d x}}}$$

Olkoon $$$u=\sqrt{x} \ln{\left(2 \right)}$$$.

Tällöin $$$du=\left(\sqrt{x} \ln{\left(2 \right)}\right)^{\prime }dx = \frac{\ln{\left(2 \right)}}{2 \sqrt{x}} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\frac{dx}{\sqrt{x}} = \frac{2 du}{\ln{\left(2 \right)}}$$$.

Siis,

$${\color{red}{\int{e^{\sqrt{x} \ln{\left(2 \right)}} d x}}} = {\color{red}{\int{\frac{2 u e^{u}}{\ln{\left(2 \right)}^{2}} d u}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{2}{\ln{\left(2 \right)}^{2}}$$$ ja $$$f{\left(u \right)} = u e^{u}$$$:

$${\color{red}{\int{\frac{2 u e^{u}}{\ln{\left(2 \right)}^{2}} d u}}} = {\color{red}{\left(\frac{2 \int{u e^{u} d u}}{\ln{\left(2 \right)}^{2}}\right)}}$$

Integraalin $$$\int{u e^{u} d u}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{\theta} \operatorname{dv} = \operatorname{\theta}\operatorname{v} - \int \operatorname{v} \operatorname{d\theta}$$$.

Olkoon $$$\operatorname{\theta}=u$$$ ja $$$\operatorname{dv}=e^{u} du$$$.

Tällöin $$$\operatorname{d\theta}=\left(u\right)^{\prime }du=1 du$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (vaiheet ovat nähtävissä »).

Näin ollen,

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

Eksponenttifunktion integraali on $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{2 \left(u e^{u} - {\color{red}{\int{e^{u} d u}}}\right)}{\ln{\left(2 \right)}^{2}} = \frac{2 \left(u e^{u} - {\color{red}{e^{u}}}\right)}{\ln{\left(2 \right)}^{2}}$$

Muista, että $$$u=\sqrt{x} \ln{\left(2 \right)}$$$:

$$\frac{2 \left(- e^{{\color{red}{u}}} + {\color{red}{u}} e^{{\color{red}{u}}}\right)}{\ln{\left(2 \right)}^{2}} = \frac{2 \left(- e^{{\color{red}{\sqrt{x} \ln{\left(2 \right)}}}} + {\color{red}{\sqrt{x} \ln{\left(2 \right)}}} e^{{\color{red}{\sqrt{x} \ln{\left(2 \right)}}}}\right)}{\ln{\left(2 \right)}^{2}}$$

Näin ollen,

$$\int{2^{\sqrt{x}} d x} = \frac{2 \left(\sqrt{x} e^{\sqrt{x} \ln{\left(2 \right)}} \ln{\left(2 \right)} - e^{\sqrt{x} \ln{\left(2 \right)}}\right)}{\ln{\left(2 \right)}^{2}}$$

Sievennä:

$$\int{2^{\sqrt{x}} d x} = \frac{2 \left(\sqrt{x} \ln{\left(2 \right)} - 1\right) e^{\sqrt{x} \ln{\left(2 \right)}}}{\ln{\left(2 \right)}^{2}}$$

Lisää integrointivakio:

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

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