Integral dari $$$\frac{x}{\left(x^{2} + 1\right)^{2}}$$$

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

Misalkan $$$u=x^{2} + 1$$$.

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

Oleh karena itu,

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

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

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

Terapkan aturan pangkat $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-2$$$:

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

Ingat bahwa $$$u=x^{2} + 1$$$:

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

Oleh karena itu,

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

Sederhanakan:

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

Tambahkan konstanta integrasi:

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

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