Integral dari $$$\frac{1}{\sqrt{x + 1} + \sqrt{x + 2}}$$$

Temukan $$$\int \frac{1}{\sqrt{x + 1} + \sqrt{x + 2}}\, dx$$$.

Rasionalkan penyebutnya:

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

Integralkan suku demi suku:

$${\color{red}{\int{\left(- \sqrt{x + 1} + \sqrt{x + 2}\right)d x}}} = {\color{red}{\left(- \int{\sqrt{x + 1} d x} + \int{\sqrt{x + 2} d x}\right)}}$$

Misalkan $$$u=x + 2$$$.

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

Integralnya menjadi

$$- \int{\sqrt{x + 1} d x} + {\color{red}{\int{\sqrt{x + 2} d x}}} = - \int{\sqrt{x + 1} d x} + {\color{red}{\int{\sqrt{u} d u}}}$$

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

$$- \int{\sqrt{x + 1} d x} + {\color{red}{\int{\sqrt{u} d u}}}=- \int{\sqrt{x + 1} d x} + {\color{red}{\int{u^{\frac{1}{2}} d u}}}=- \int{\sqrt{x + 1} d x} + {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- \int{\sqrt{x + 1} d x} + {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Ingat bahwa $$$u=x + 2$$$:

$$- \int{\sqrt{x + 1} d x} + \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = - \int{\sqrt{x + 1} d x} + \frac{2 {\color{red}{\left(x + 2\right)}}^{\frac{3}{2}}}{3}$$

Misalkan $$$u=x + 1$$$.

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

Integral tersebut dapat ditulis ulang sebagai

$$\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\int{\sqrt{x + 1} d x}}} = \frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\int{\sqrt{u} d u}}}$$

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

$$\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\int{\sqrt{u} d u}}}=\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\int{u^{\frac{1}{2}} d u}}}=\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Ingat bahwa $$$u=x + 1$$$:

$$\frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = \frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3} - \frac{2 {\color{red}{\left(x + 1\right)}}^{\frac{3}{2}}}{3}$$

Oleh karena itu,

$$\int{\frac{1}{\sqrt{x + 1} + \sqrt{x + 2}} d x} = - \frac{2 \left(x + 1\right)^{\frac{3}{2}}}{3} + \frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3}$$

Tambahkan konstanta integrasi:

$$\int{\frac{1}{\sqrt{x + 1} + \sqrt{x + 2}} d x} = - \frac{2 \left(x + 1\right)^{\frac{3}{2}}}{3} + \frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3}+C$$

$$$\int \frac{1}{\sqrt{x + 1} + \sqrt{x + 2}}\, dx = \left(- \frac{2 \left(x + 1\right)^{\frac{3}{2}}}{3} + \frac{2 \left(x + 2\right)^{\frac{3}{2}}}{3}\right) + C$$$A