Integral of $$$\left(a t - b t\right)^{2}$$$ with respect to $$$t$$$

Find $$$\int \left(a t - b t\right)^{2}\, dt$$$.

Let $$$u=a t - b t$$$.

Then $$$du=\left(a t - b t\right)^{\prime }dt = \left(a - b\right) dt$$$ (steps can be seen »), and we have that $$$dt = \frac{du}{a - b}$$$.

Therefore,

$${\color{red}{\int{\left(a t - b t\right)^{2} d t}}} = {\color{red}{\int{\frac{u^{2}}{a - b} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{a - b}$$$ and $$$f{\left(u \right)} = u^{2}$$$:

$${\color{red}{\int{\frac{u^{2}}{a - b} d u}}} = {\color{red}{\frac{\int{u^{2} d u}}{a - b}}}$$

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

$$\frac{{\color{red}{\int{u^{2} d u}}}}{a - b}=\frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{a - b}=\frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{a - b}$$

Recall that $$$u=a t - b t$$$:

$$\frac{{\color{red}{u}}^{3}}{3 \left(a - b\right)} = \frac{{\color{red}{\left(a t - b t\right)}}^{3}}{3 \left(a - b\right)}$$

Therefore,

$$\int{\left(a t - b t\right)^{2} d t} = \frac{\left(a t - b t\right)^{3}}{3 \left(a - b\right)}$$

Simplify:

$$\int{\left(a t - b t\right)^{2} d t} = \frac{t^{3} \left(- a + b\right)^{2}}{3}$$

Add the constant of integration:

$$\int{\left(a t - b t\right)^{2} d t} = \frac{t^{3} \left(- a + b\right)^{2}}{3}+C$$

$$$\int \left(a t - b t\right)^{2}\, dt = \frac{t^{3} \left(- a + b\right)^{2}}{3} + C$$$A