Integral dari $$$9 e^{\sqrt{x}}$$$

Temukan $$$\int 9 e^{\sqrt{x}}\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=9$$$ dan $$$f{\left(x \right)} = e^{\sqrt{x}}$$$:

$${\color{red}{\int{9 e^{\sqrt{x}} d x}}} = {\color{red}{\left(9 \int{e^{\sqrt{x}} d x}\right)}}$$

Misalkan $$$u=\sqrt{x}$$$.

Kemudian $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

Jadi,

$$9 {\color{red}{\int{e^{\sqrt{x}} d x}}} = 9 {\color{red}{\int{2 u e^{u} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=2$$$ dan $$$f{\left(u \right)} = u e^{u}$$$:

$$9 {\color{red}{\int{2 u e^{u} d u}}} = 9 {\color{red}{\left(2 \int{u e^{u} d u}\right)}}$$

Untuk integral $$$\int{u e^{u} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{\mu} \operatorname{dv} = \operatorname{\mu}\operatorname{v} - \int \operatorname{v} \operatorname{d\mu}$$$.

Misalkan $$$\operatorname{\mu}=u$$$ dan $$$\operatorname{dv}=e^{u} du$$$.

Maka $$$\operatorname{d\mu}=\left(u\right)^{\prime }du=1 du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (langkah-langkah dapat dilihat di »).

Oleh karena itu,

$$18 {\color{red}{\int{u e^{u} d u}}}=18 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=18 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$

Integral dari fungsi eksponensial adalah $$$\int{e^{u} d u} = e^{u}$$$:

$$18 u e^{u} - 18 {\color{red}{\int{e^{u} d u}}} = 18 u e^{u} - 18 {\color{red}{e^{u}}}$$

Ingat bahwa $$$u=\sqrt{x}$$$:

$$- 18 e^{{\color{red}{u}}} + 18 {\color{red}{u}} e^{{\color{red}{u}}} = - 18 e^{{\color{red}{\sqrt{x}}}} + 18 {\color{red}{\sqrt{x}}} e^{{\color{red}{\sqrt{x}}}}$$

Oleh karena itu,

$$\int{9 e^{\sqrt{x}} d x} = 18 \sqrt{x} e^{\sqrt{x}} - 18 e^{\sqrt{x}}$$

Sederhanakan:

$$\int{9 e^{\sqrt{x}} d x} = 18 \left(\sqrt{x} - 1\right) e^{\sqrt{x}}$$

Tambahkan konstanta integrasi:

$$\int{9 e^{\sqrt{x}} d x} = 18 \left(\sqrt{x} - 1\right) e^{\sqrt{x}}+C$$

$$$\int 9 e^{\sqrt{x}}\, dx = 18 \left(\sqrt{x} - 1\right) e^{\sqrt{x}} + C$$$A