$$$\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln\left(x + \sqrt{x^{2} + 1}\right)}{2}$$$'nin integrali

Bulun: $$$\int \left(\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln\left(x + \sqrt{x^{2} + 1}\right)}{2}\right)\, dx$$$.

Her terimin integralini alın:

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

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

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

$$$\int{\ln{\left(x + \sqrt{x^{2} + 1} \right)} d x}$$$ integrali için, kısmi integrasyonu $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ kullanın.

$$$\operatorname{u}=\ln{\left(x + \sqrt{x^{2} + 1} \right)}$$$ ve $$$\operatorname{dv}=dx$$$ olsun.

O halde $$$\operatorname{du}=\left(\ln{\left(x + \sqrt{x^{2} + 1} \right)}\right)^{\prime }dx=\frac{dx}{\sqrt{x^{2} + 1}}$$$ (adımlar için bkz. ») ve $$$\operatorname{v}=\int{1 d x}=x$$$ (adımlar için bkz. »).

Dolayısıyla,

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

$$$u=x^{2} + 1$$$ olsun.

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

O halde,

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

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:

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

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

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

Hatırlayın ki $$$u=x^{2} + 1$$$:

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

$$$u=x^{2} + 1$$$ olsun.

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

O halde,

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{4}$$$ ve $$$f{\left(u \right)} = \sqrt{u}$$$ ile uygula:

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

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

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

Hatırlayın ki $$$u=x^{2} + 1$$$:

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

Dolayısıyla,

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

İntegrasyon sabitini ekleyin:

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

$$$\int \left(\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln\left(x + \sqrt{x^{2} + 1}\right)}{2}\right)\, dx = \left(\frac{x \ln\left(x + \sqrt{x^{2} + 1}\right)}{2} + \frac{\left(x^{2} + 1\right)^{\frac{3}{2}}}{6} - \frac{\sqrt{x^{2} + 1}}{2}\right) + C$$$A