Correct me if I'm laying out the question wrong.
We want to evaluate the following:
[tex]\int(et-4)dt[/tex]
To solve, let the following be true:
[tex]u=et-4[/tex]
This implies:
[tex]du= \frac{d}{dt}(et-4)=e(dt) [/tex]
From the previous, we solve for [tex]t[/tex] and [tex]dt[/tex] in terms of [tex]u[/tex] and [tex]du[/tex]:
[tex]t= \frac{u+4}{e} [/tex]
[tex]dt= \frac{du}{e} [/tex]
Now that we have [tex]t[/tex] and [tex]dt[/tex] in terms of [tex]u[/tex] and [tex]du[/tex] we rewrite our original integral -also- in terms of [tex]u[/tex] and [tex]du[/tex]:
[tex]\int(et-4)dt=\int[(e(\frac{u+4}{e})-4)(\frac{du}{e})]= \frac{1}{e} \int(u)du[/tex]
We solve it:
[tex] \frac{1}{e} \int(u)du= \frac{u^2}{2e}+C [/tex]
Which in terms of [tex]t[/tex] is written as:
[tex] \frac{(et-4)^2}{2e}+C [/tex]
Which we can expand to be:
[tex]\frac{e^2t^2-8et+16}{2e}+C= \frac{et^2}{2} - 4t+ \frac{16}{2e} +C[/tex]
We can incorporate [tex] \frac{16}{2e} [/tex] to the integration primitive [tex]C[/tex], as it is also a constant.
Hence, the final form of the integral is:
[tex]\frac{et^2}{2} - 4t+C[/tex]
