Integral dari $$$\sqrt{\frac{x}{1 - x}}$$$

Kalkulator akan menemukan integral/antiturunan dari $$$\sqrt{\frac{x}{1 - x}}$$$, dengan menampilkan langkah-langkah.

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Masukan Anda

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

Solusi

Masukan ditulis ulang: $$$\int{\sqrt{\frac{x}{1 - x}} d x}=\int{\frac{\sqrt{x}}{\sqrt{1 - x}} d x}$$$.

Misalkan $$$u=\sqrt{x}$$$.

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

Integralnya menjadi

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

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

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

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

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

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

Dengan demikian,

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

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

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

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

$$$\frac{\sin^{2}{\left( v \right)}}{\sqrt{\cos^{2}{\left( v \right)}}} = \frac{\sin^{2}{\left( v \right)}}{\cos{\left( v \right)}}$$$

Oleh karena itu,

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

Terapkan rumus reduksi pangkat $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ dengan $$$\alpha= v $$$:

$$2 {\color{red}{\int{\sin^{2}{\left(v \right)} d v}}} = 2 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 v \right)}}{2}\right)d v}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(v \right)} = 1 - \cos{\left(2 v \right)}$$$:

$$2 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 v \right)}}{2}\right)d v}}} = 2 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 v \right)}\right)d v}}{2}\right)}}$$

Integralkan suku demi suku:

$${\color{red}{\int{\left(1 - \cos{\left(2 v \right)}\right)d v}}} = {\color{red}{\left(\int{1 d v} - \int{\cos{\left(2 v \right)} d v}\right)}}$$

Terapkan aturan konstanta $$$\int c\, dv = c v$$$ dengan $$$c=1$$$:

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

Misalkan $$$w=2 v$$$.

Kemudian $$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dv = \frac{dw}{2}$$$.

Integralnya menjadi

$$v - {\color{red}{\int{\cos{\left(2 v \right)} d v}}} = v - {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(w \right)} = \cos{\left(w \right)}$$$:

$$v - {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}} = v - {\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{2}\right)}}$$

Integral dari kosinus adalah $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$:

$$v - \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{2} = v - \frac{{\color{red}{\sin{\left(w \right)}}}}{2}$$

Ingat bahwa $$$w=2 v$$$:

$$v - \frac{\sin{\left({\color{red}{w}} \right)}}{2} = v - \frac{\sin{\left({\color{red}{\left(2 v\right)}} \right)}}{2}$$

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

$$- \frac{\sin{\left(2 {\color{red}{v}} \right)}}{2} + {\color{red}{v}} = - \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(u \right)}}} \right)}}{2} + {\color{red}{\operatorname{asin}{\left(u \right)}}}$$

Ingat bahwa $$$u=\sqrt{x}$$$:

$$- \frac{\sin{\left(2 \operatorname{asin}{\left({\color{red}{u}} \right)} \right)}}{2} + \operatorname{asin}{\left({\color{red}{u}} \right)} = - \frac{\sin{\left(2 \operatorname{asin}{\left({\color{red}{\sqrt{x}}} \right)} \right)}}{2} + \operatorname{asin}{\left({\color{red}{\sqrt{x}}} \right)}$$

Oleh karena itu,

$$\int{\frac{\sqrt{x}}{\sqrt{1 - x}} d x} = - \frac{\sin{\left(2 \operatorname{asin}{\left(\sqrt{x} \right)} \right)}}{2} + \operatorname{asin}{\left(\sqrt{x} \right)}$$

Dengan menggunakan rumus $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, sederhanakan ekspresi:

$$\int{\frac{\sqrt{x}}{\sqrt{1 - x}} d x} = - \sqrt{x} \sqrt{1 - x} + \operatorname{asin}{\left(\sqrt{x} \right)}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sqrt{x}}{\sqrt{1 - x}} d x} = - \sqrt{x} \sqrt{1 - x} + \operatorname{asin}{\left(\sqrt{x} \right)}+C$$

Jawaban

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