I am solving the following problem using $u$-substitution:
$$\int_0^1 \frac{t+1}{2t^2+4t+3}dt$$
So $u$ = $2t^2+4t+3$ and $du$ = $4(t+1)dt$. However this does not exist in the equation so it must be balanced out. I am fairly certain that the problem ends up being something like this:
$$\int_3^9 \frac{1}{u} \frac{du}{4}$$ but I get confused as to how exactly the $t+1$ on top gets cancelled out.
The $t + 1$ goes away when you make the substitution $dt = \frac{du}{4(t+1)}$ in your integrand. So,
$$ \int_0^1\frac{t+1}{2t^2+4t+3}\,dt = \int_3^9\frac{t+1}{u}\frac{du}{4(t+1)} = \int_3^9\frac{du}{4u}. $$