$$$\frac{t - 4}{\sqrt{t}}$$$'nin integrali

Bulun: $$$\int \frac{t - 4}{\sqrt{t}}\, dt$$$.

Expand the expression:

$${\color{red}{\int{\frac{t - 4}{\sqrt{t}} d t}}} = {\color{red}{\int{\left(\sqrt{t} - \frac{4}{\sqrt{t}}\right)d t}}}$$

Her terimin integralini alın:

$${\color{red}{\int{\left(\sqrt{t} - \frac{4}{\sqrt{t}}\right)d t}}} = {\color{red}{\left(- \int{\frac{4}{\sqrt{t}} d t} + \int{\sqrt{t} d t}\right)}}$$

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

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

Sabit katsayı kuralı $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$'i $$$c=4$$$ ve $$$f{\left(t \right)} = \frac{1}{\sqrt{t}}$$$ ile uygula:

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

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

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

Dolayısıyla,

$$\int{\frac{t - 4}{\sqrt{t}} d t} = \frac{2 t^{\frac{3}{2}}}{3} - 8 \sqrt{t}$$

Sadeleştirin:

$$\int{\frac{t - 4}{\sqrt{t}} d t} = \frac{2 \sqrt{t} \left(t - 12\right)}{3}$$

İntegrasyon sabitini ekleyin:

$$\int{\frac{t - 4}{\sqrt{t}} d t} = \frac{2 \sqrt{t} \left(t - 12\right)}{3}+C$$

$$$\int \frac{t - 4}{\sqrt{t}}\, dt = \frac{2 \sqrt{t} \left(t - 12\right)}{3} + C$$$A