$$$\frac{x + 2}{\sqrt{2 x + 1}}$$$'nin integrali

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

Payı $$$x + 2=\frac{2 x + 1}{2} + \frac{3}{2}$$$ olarak yeniden yazın ve kesri ayırın.:

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

Her terimin integralini alın:

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

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \sqrt{2 x + 1}$$$ ile uygula:

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

$$$u=2 x + 1$$$ olsun.

Böylece $$$du=\left(2 x + 1\right)^{\prime }dx = 2 dx$$$ (adımlar » görülebilir) ve $$$dx = \frac{du}{2}$$$ elde ederiz.

Dolayısıyla,

$$\int{\frac{3}{2 \sqrt{2 x + 1}} d x} + \frac{{\color{red}{\int{\sqrt{2 x + 1} d x}}}}{2} = \int{\frac{3}{2 \sqrt{2 x + 1}} d x} + \frac{{\color{red}{\int{\frac{\sqrt{u}}{2} d u}}}}{2}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(u \right)} = \sqrt{u}$$$ ile uygula:

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

Kuvvet kuralını $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=\frac{1}{2}$$$ ile uygulayın:

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

Hatırlayın ki $$$u=2 x + 1$$$:

$$\int{\frac{3}{2 \sqrt{2 x + 1}} d x} + \frac{{\color{red}{u}}^{\frac{3}{2}}}{6} = \int{\frac{3}{2 \sqrt{2 x + 1}} d x} + \frac{{\color{red}{\left(2 x + 1\right)}}^{\frac{3}{2}}}{6}$$

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{3}{2}$$$ ve $$$f{\left(x \right)} = \frac{1}{\sqrt{2 x + 1}}$$$ ile uygula:

$$\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + {\color{red}{\int{\frac{3}{2 \sqrt{2 x + 1}} d x}}} = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + {\color{red}{\left(\frac{3 \int{\frac{1}{\sqrt{2 x + 1}} d x}}{2}\right)}}$$

$$$u=2 x + 1$$$ olsun.

Böylece $$$du=\left(2 x + 1\right)^{\prime }dx = 2 dx$$$ (adımlar » görülebilir) ve $$$dx = \frac{du}{2}$$$ elde ederiz.

İntegral şu hale gelir

$$\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\int{\frac{1}{\sqrt{2 x + 1}} d x}}}}{2} = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}}}{2}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ ile uygula:

$$\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}}}{2} = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{u}} d u}}{2}\right)}}}{2}$$

Kuvvet kuralını $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=- \frac{1}{2}$$$ ile uygulayın:

$$\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{4}=\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{4}=\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{4}=\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{4}=\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 {\color{red}{\left(2 \sqrt{u}\right)}}}{4}$$

Hatırlayın ki $$$u=2 x + 1$$$:

$$\frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 \sqrt{{\color{red}{u}}}}{2} = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 \sqrt{{\color{red}{\left(2 x + 1\right)}}}}{2}$$

Dolayısıyla,

$$\int{\frac{x + 2}{\sqrt{2 x + 1}} d x} = \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} + \frac{3 \sqrt{2 x + 1}}{2}$$

Sadeleştirin:

$$\int{\frac{x + 2}{\sqrt{2 x + 1}} d x} = \frac{\left(x + 5\right) \sqrt{2 x + 1}}{3}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{x + 2}{\sqrt{2 x + 1}} d x} = \frac{\left(x + 5\right) \sqrt{2 x + 1}}{3}+C$$

$$$\int \frac{x + 2}{\sqrt{2 x + 1}}\, dx = \frac{\left(x + 5\right) \sqrt{2 x + 1}}{3} + C$$$A