Integral of $$$z \left(1 + \frac{z}{t}\right)$$$ with respect to $$$z$$$
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Find $$$\int z \left(1 + \frac{z}{t}\right)\, dz$$$.
Solution
Expand the expression:
$${\color{red}{\int{z \left(1 + \frac{z}{t}\right) d z}}} = {\color{red}{\int{\left(z + \frac{z^{2}}{t}\right)d z}}}$$
Integrate term by term:
$${\color{red}{\int{\left(z + \frac{z^{2}}{t}\right)d z}}} = {\color{red}{\left(\int{z d z} + \int{\frac{z^{2}}{t} d z}\right)}}$$
Apply the power rule $$$\int z^{n}\, dz = \frac{z^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\int{\frac{z^{2}}{t} d z} + {\color{red}{\int{z d z}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\frac{z^{1 + 1}}{1 + 1}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\left(\frac{z^{2}}{2}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(z \right)}\, dz = c \int f{\left(z \right)}\, dz$$$ with $$$c=\frac{1}{t}$$$ and $$$f{\left(z \right)} = z^{2}$$$:
$$\frac{z^{2}}{2} + {\color{red}{\int{\frac{z^{2}}{t} d z}}} = \frac{z^{2}}{2} + {\color{red}{\frac{\int{z^{2} d z}}{t}}}$$
Apply the power rule $$$\int z^{n}\, dz = \frac{z^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$\frac{z^{2}}{2} + \frac{{\color{red}{\int{z^{2} d z}}}}{t}=\frac{z^{2}}{2} + \frac{{\color{red}{\frac{z^{1 + 2}}{1 + 2}}}}{t}=\frac{z^{2}}{2} + \frac{{\color{red}{\left(\frac{z^{3}}{3}\right)}}}{t}$$
Therefore,
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2}}{2} + \frac{z^{3}}{3 t}$$
Simplify:
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}$$
Add the constant of integration:
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}+C$$
Answer: $$$\int{z \left(1 + \frac{z}{t}\right) d z}=\frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}+C$$$