Integral of $$$-1 + \frac{\sqrt{x}}{t}$$$ with respect to $$$x$$$

Find $$$\int \left(-1 + \frac{\sqrt{x}}{t}\right)\, dx$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$\int{\frac{\sqrt{x}}{t} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{\sqrt{x}}{t} d x} - {\color{red}{x}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{t}$$$ and $$$f{\left(x \right)} = \sqrt{x}$$$:

$$- x + {\color{red}{\int{\frac{\sqrt{x}}{t} d x}}} = - x + {\color{red}{\frac{\int{\sqrt{x} d x}}{t}}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

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

Therefore,

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

Add the constant of integration:

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

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