Integral dari $$$\frac{a^{\sqrt{x}}}{\sqrt{x}}$$$ terhadap $$$x$$$

Temukan $$$\int \frac{a^{\sqrt{x}}}{\sqrt{x}}\, dx$$$.

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

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

Jadi,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{2}{\ln{\left(a \right)}}$$$ dan $$$f{\left(u \right)} = 1$$$:

$${\color{red}{\int{\frac{2}{\ln{\left(a \right)}} d u}}} = {\color{red}{\left(\frac{2 \int{1 d u}}{\ln{\left(a \right)}}\right)}}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

$$\frac{2 {\color{red}{\int{1 d u}}}}{\ln{\left(a \right)}} = \frac{2 {\color{red}{u}}}{\ln{\left(a \right)}}$$

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

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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