Integral dari $$$\frac{\sqrt{y^{5} - 1}}{y}$$$

Temukan $$$\int \frac{\sqrt{y^{5} - 1}}{y}\, dy$$$.

Misalkan $$$u=y^{\frac{5}{2}}$$$.

Kemudian $$$du=\left(y^{\frac{5}{2}}\right)^{\prime }dy = \frac{5 y^{\frac{3}{2}}}{2} dy$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$y^{\frac{3}{2}} dy = \frac{2 du}{5}$$$.

Dengan demikian,

$${\color{red}{\int{\frac{\sqrt{y^{5} - 1}}{y} d y}}} = {\color{red}{\int{\frac{2 \sqrt{u^{2} - 1}}{5 u} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{2}{5}$$$ dan $$$f{\left(u \right)} = \frac{\sqrt{u^{2} - 1}}{u}$$$:

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

Misalkan $$$u=\cosh{\left(v \right)}$$$.

Maka $$$du=\left(\cosh{\left(v \right)}\right)^{\prime }dv = \sinh{\left(v \right)} dv$$$ (langkah-langkah dapat dilihat »).

Selain itu, berlaku $$$v=\operatorname{acosh}{\left(u \right)}$$$.

Integran menjadi

$$$\frac{\sqrt{ u ^{2} - 1}}{ u } = \frac{\sqrt{\cosh^{2}{\left( v \right)} - 1}}{\cosh{\left( v \right)}}$$$

Gunakan identitas $$$\cosh^{2}{\left( v \right)} - 1 = \sinh^{2}{\left( v \right)}$$$:

$$$\frac{\sqrt{\cosh^{2}{\left( v \right)} - 1}}{\cosh{\left( v \right)}}=\frac{\sqrt{\sinh^{2}{\left( v \right)}}}{\cosh{\left( v \right)}}$$$

Dengan asumsi bahwa $$$\sinh{\left( v \right)} \ge 0$$$, diperoleh sebagai berikut:

$$$\frac{\sqrt{\sinh^{2}{\left( v \right)}}}{\cosh{\left( v \right)}} = \frac{\sinh{\left( v \right)}}{\cosh{\left( v \right)}}$$$

Dengan demikian,

$$\frac{2 {\color{red}{\int{\frac{\sqrt{u^{2} - 1}}{u} d u}}}}{5} = \frac{2 {\color{red}{\int{\frac{\sinh^{2}{\left(v \right)}}{\cosh{\left(v \right)}} d v}}}}{5}$$

Kalikan pembilang dan penyebut dengan satu faktor kosinus hiperbolik dan tuliskan sisanya dalam bentuk sinus hiperbolik, menggunakan rumus $$$\cosh^2\left(\alpha \right)=\sinh^2\left(\alpha \right)+1$$$ dengan $$$\alpha= v $$$:

$$\frac{2 {\color{red}{\int{\frac{\sinh^{2}{\left(v \right)}}{\cosh{\left(v \right)}} d v}}}}{5} = \frac{2 {\color{red}{\int{\frac{\sinh^{2}{\left(v \right)} \cosh{\left(v \right)}}{\sinh^{2}{\left(v \right)} + 1} d v}}}}{5}$$

Misalkan $$$w=\sinh{\left(v \right)}$$$.

Kemudian $$$dw=\left(\sinh{\left(v \right)}\right)^{\prime }dv = \cosh{\left(v \right)} dv$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\cosh{\left(v \right)} dv = dw$$$.

Oleh karena itu,

$$\frac{2 {\color{red}{\int{\frac{\sinh^{2}{\left(v \right)} \cosh{\left(v \right)}}{\sinh^{2}{\left(v \right)} + 1} d v}}}}{5} = \frac{2 {\color{red}{\int{\frac{w^{2}}{w^{2} + 1} d w}}}}{5}$$

Tulis ulang dan pisahkan pecahannya:

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

Integralkan suku demi suku:

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

Terapkan aturan konstanta $$$\int c\, dw = c w$$$ dengan $$$c=1$$$:

$$- \frac{2 \int{\frac{1}{w^{2} + 1} d w}}{5} + \frac{2 {\color{red}{\int{1 d w}}}}{5} = - \frac{2 \int{\frac{1}{w^{2} + 1} d w}}{5} + \frac{2 {\color{red}{w}}}{5}$$

Integral dari $$$\frac{1}{w^{2} + 1}$$$ adalah $$$\int{\frac{1}{w^{2} + 1} d w} = \operatorname{atan}{\left(w \right)}$$$:

$$\frac{2 w}{5} - \frac{2 {\color{red}{\int{\frac{1}{w^{2} + 1} d w}}}}{5} = \frac{2 w}{5} - \frac{2 {\color{red}{\operatorname{atan}{\left(w \right)}}}}{5}$$

Ingat bahwa $$$w=\sinh{\left(v \right)}$$$:

$$- \frac{2 \operatorname{atan}{\left({\color{red}{w}} \right)}}{5} + \frac{2 {\color{red}{w}}}{5} = - \frac{2 \operatorname{atan}{\left({\color{red}{\sinh{\left(v \right)}}} \right)}}{5} + \frac{2 {\color{red}{\sinh{\left(v \right)}}}}{5}$$

Ingat bahwa $$$v=\operatorname{acosh}{\left(u \right)}$$$:

$$\frac{2 \sinh{\left({\color{red}{v}} \right)}}{5} - \frac{2 \operatorname{atan}{\left(\sinh{\left({\color{red}{v}} \right)} \right)}}{5} = \frac{2 \sinh{\left({\color{red}{\operatorname{acosh}{\left(u \right)}}} \right)}}{5} - \frac{2 \operatorname{atan}{\left(\sinh{\left({\color{red}{\operatorname{acosh}{\left(u \right)}}} \right)} \right)}}{5}$$

Ingat bahwa $$$u=y^{\frac{5}{2}}$$$:

$$\frac{2 \sqrt{1 + {\color{red}{u}}} \sqrt{-1 + {\color{red}{u}}}}{5} - \frac{2 \operatorname{atan}{\left(\sqrt{1 + {\color{red}{u}}} \sqrt{-1 + {\color{red}{u}}} \right)}}{5} = \frac{2 \sqrt{1 + {\color{red}{y^{\frac{5}{2}}}}} \sqrt{-1 + {\color{red}{y^{\frac{5}{2}}}}}}{5} - \frac{2 \operatorname{atan}{\left(\sqrt{1 + {\color{red}{y^{\frac{5}{2}}}}} \sqrt{-1 + {\color{red}{y^{\frac{5}{2}}}}} \right)}}{5}$$

Oleh karena itu,

$$\int{\frac{\sqrt{y^{5} - 1}}{y} d y} = \frac{2 \sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1}}{5} - \frac{2 \operatorname{atan}{\left(\sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1} \right)}}{5}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sqrt{y^{5} - 1}}{y} d y} = \frac{2 \sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1}}{5} - \frac{2 \operatorname{atan}{\left(\sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1} \right)}}{5}+C$$

$$$\int \frac{\sqrt{y^{5} - 1}}{y}\, dy = \left(\frac{2 \sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1}}{5} - \frac{2 \operatorname{atan}{\left(\sqrt{y^{\frac{5}{2}} - 1} \sqrt{y^{\frac{5}{2}} + 1} \right)}}{5}\right) + C$$$A