Integral dari $$$\frac{\sqrt{x}}{x^{\frac{3}{4}} + 1}$$$

Temukan $$$\int \frac{\sqrt{x}}{x^{\frac{3}{4}} + 1}\, dx$$$.

Misalkan $$$u=\sqrt[4]{x}$$$.

Kemudian $$$du=\left(\sqrt[4]{x}\right)^{\prime }dx = \frac{1}{4 x^{\frac{3}{4}}} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\frac{dx}{x^{\frac{3}{4}}} = 4 du$$$.

Integralnya menjadi

$${\color{red}{\int{\frac{\sqrt{x}}{x^{\frac{3}{4}} + 1} d x}}} = {\color{red}{\int{\frac{4 u^{5}}{u^{3} + 1} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=4$$$ dan $$$f{\left(u \right)} = \frac{u^{5}}{u^{3} + 1}$$$:

$${\color{red}{\int{\frac{4 u^{5}}{u^{3} + 1} d u}}} = {\color{red}{\left(4 \int{\frac{u^{5}}{u^{3} + 1} d u}\right)}}$$

Karena derajat pembilang tidak kurang dari derajat penyebut, lakukan pembagian panjang polinom (langkah-langkah dapat dilihat »):

$$4 {\color{red}{\int{\frac{u^{5}}{u^{3} + 1} d u}}} = 4 {\color{red}{\int{\left(u^{2} - \frac{u^{2}}{u^{3} + 1}\right)d u}}}$$

Integralkan suku demi suku:

$$4 {\color{red}{\int{\left(u^{2} - \frac{u^{2}}{u^{3} + 1}\right)d u}}} = 4 {\color{red}{\left(\int{u^{2} d u} - \int{\frac{u^{2}}{u^{3} + 1} d u}\right)}}$$

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

$$- 4 \int{\frac{u^{2}}{u^{3} + 1} d u} + 4 {\color{red}{\int{u^{2} d u}}}=- 4 \int{\frac{u^{2}}{u^{3} + 1} d u} + 4 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 4 \int{\frac{u^{2}}{u^{3} + 1} d u} + 4 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Misalkan $$$v=u^{3} + 1$$$.

Kemudian $$$dv=\left(u^{3} + 1\right)^{\prime }du = 3 u^{2} du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$u^{2} du = \frac{dv}{3}$$$.

Jadi,

$$\frac{4 u^{3}}{3} - 4 {\color{red}{\int{\frac{u^{2}}{u^{3} + 1} d u}}} = \frac{4 u^{3}}{3} - 4 {\color{red}{\int{\frac{1}{3 v} d v}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{3}$$$ dan $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$\frac{4 u^{3}}{3} - 4 {\color{red}{\int{\frac{1}{3 v} d v}}} = \frac{4 u^{3}}{3} - 4 {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{3}\right)}}$$

Integral dari $$$\frac{1}{v}$$$ adalah $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{4 u^{3}}{3} - \frac{4 {\color{red}{\int{\frac{1}{v} d v}}}}{3} = \frac{4 u^{3}}{3} - \frac{4 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{3}$$

Ingat bahwa $$$v=u^{3} + 1$$$:

$$\frac{4 u^{3}}{3} - \frac{4 \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{3} = \frac{4 u^{3}}{3} - \frac{4 \ln{\left(\left|{{\color{red}{\left(u^{3} + 1\right)}}}\right| \right)}}{3}$$

Ingat bahwa $$$u=\sqrt[4]{x}$$$:

$$- \frac{4 \ln{\left(\left|{1 + {\color{red}{u}}^{3}}\right| \right)}}{3} + \frac{4 {\color{red}{u}}^{3}}{3} = - \frac{4 \ln{\left(\left|{1 + {\color{red}{\sqrt[4]{x}}}^{3}}\right| \right)}}{3} + \frac{4 {\color{red}{\sqrt[4]{x}}}^{3}}{3}$$

Oleh karena itu,

$$\int{\frac{\sqrt{x}}{x^{\frac{3}{4}} + 1} d x} = \frac{4 x^{\frac{3}{4}}}{3} - \frac{4 \ln{\left(\left|{x^{\frac{3}{4}} + 1}\right| \right)}}{3}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sqrt{x}}{x^{\frac{3}{4}} + 1} d x} = \frac{4 x^{\frac{3}{4}}}{3} - \frac{4 \ln{\left(\left|{x^{\frac{3}{4}} + 1}\right| \right)}}{3}+C$$

$$$\int \frac{\sqrt{x}}{x^{\frac{3}{4}} + 1}\, dx = \left(\frac{4 x^{\frac{3}{4}}}{3} - \frac{4 \ln\left(\left|{x^{\frac{3}{4}} + 1}\right|\right)}{3}\right) + C$$$A