Integral dari $$$\frac{t^{2} e^{- x^{2}}}{u}$$$ terhadap $$$x$$$

Temukan $$$\int \frac{t^{2} e^{- x^{2}}}{u}\, dx$$$.

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

$${\color{red}{\int{\frac{t^{2} e^{- x^{2}}}{u} d x}}} = {\color{red}{\frac{t^{2} \int{e^{- x^{2}} d x}}{u}}}$$

Integral ini (Fungsi galat) tidak memiliki bentuk tertutup:

$$\frac{t^{2} {\color{red}{\int{e^{- x^{2}} d x}}}}{u} = \frac{t^{2} {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erf}{\left(x \right)}}{2}\right)}}}{u}$$

Oleh karena itu,

$$\int{\frac{t^{2} e^{- x^{2}}}{u} d x} = \frac{\sqrt{\pi} t^{2} \operatorname{erf}{\left(x \right)}}{2 u}$$

Tambahkan konstanta integrasi:

$$\int{\frac{t^{2} e^{- x^{2}}}{u} d x} = \frac{\sqrt{\pi} t^{2} \operatorname{erf}{\left(x \right)}}{2 u}+C$$

$$$\int \frac{t^{2} e^{- x^{2}}}{u}\, dx = \frac{\sqrt{\pi} t^{2} \operatorname{erf}{\left(x \right)}}{2 u} + C$$$A