So for my homework I've gotten an incorrect answer on this problem 3 times in a row. Here's an overview of my work
A large tank holds 250 liters of water with a salt concentration of 7 grams per liter. A brine solution containing 3 grams per liter is added to the tank at a rate of 9 liters per minute. The well-mixed solution is pumped out of the tank at a rate of 5 liters per minute.
How much salt is in the tank after 15 minutes?
Enter your answer to the nearest 0.0001 grams.
$$\begin{align*}
S(t)&= \text{concentration of salt as a function of time.}\\
S'&=27-\left(\frac{5S}{250+4t}\right)\\
I&=(250+4t)^5\\
[S(250+4t)^5]'&=27(250+4t)^5\\
S(250+4t)^5&=\frac{9}{8}(250+4t)^6+C\\
S&=\frac{\frac{9}{8}(250+4t)+C}{(250+4t)^5}\\
S(0)&=1750=\frac{\frac{9}{8}(250)+C}{(250)^5}\\
C&=1468.75(250)^5\\
S(15)&=\frac{\frac{9}{8}(250+4(15))+1468.75(250)^5}{(250+4(15)^5}\\
S(15)&=849.7520
\end{align*}$$
Best Answer
This is a pretty standard "mixing problem." You went wrong in a couple of places:
You start pretty well: if we let $S(t)$ be the amount of salt (in grams) in the tank at time $t$ ($t$ measured in minutes). (Note, $S(t)$ is the amount, not the concentration; your formulas clearly view $S$ as the amount, not the concentration; see Pickahu's set-up if you want to use the concentration instead).
In these problems, the amount of salt at any given time is changing by the formula $$\frac{dS}{dt} = \binom{\text{rate}}{\text{in}} - \binom{\text{rate}}{\text{out}}.$$ And the initial condition $S(0)$ depends on the problem.
The initial condition is simple enough: you are told there are 250 liters of water, with a salt concentration of 7 grams per liter. So $$S(0) = \left(250\ \text{liters}\right)\left(7\ \frac{\text{grams}}{\text{liter}}\right) = 1750\ \text{grams of salt.}$$
What about the rates in and out? We are adding 9 liters per minute, each liter containing 3 grams of salt. That is, the rate in is: $$\text{rate in} = \left(3\frac{\text{grams}}{\text{liter}}\right)\left(9\frac{\text{liters}}{\text{minute}}\right) = 27\frac{\text{grams}}{\text{minute}};$$
What is the rate out? We are letting out 5 liters per minute; each liter will have as much salt as the concentration at time $t$. The concentration at time $t$ is given by the amount of salt at time $t$, which is $S(t)$, divided by the amount of liquid at time $t$.
From the moment we start with $250$ liters, each minute you add $9$ liters and you drain $5$ liters, for a net total addition of $4$ liters per minute. So at time $t$, the total amount of liquid in the tank is $250+4t$. So the concentration of salt at time $t$ is $$\frac{S(t)}{250+4t}\ \frac{\text{grams}}{\text{liter}}.$$
Since we are draining five liters at this concentration, we have that $$\text{rate out} = \left(5\ \frac{\text{liters}}{\text{minute}}\right)\left(\frac{S(t)}{250+4t}\ \frac{\text{grams}}{\text{liter}}\right) = \frac{5S(t)}{250+4t}\ \frac{\text{grams}}{\text{minute}}.$$
So the differential equation we need to solve is: $$\frac{dS}{dt} = 27 - \frac{5S}{250+4t}.$$
Writing this in the standard form, we have $$S' + \frac{5}{250+4t}S = 27.$$ We need an integrating factor. Letting $\mu(t)$ stand for this factor, multiplying through we have $$\mu(t)S' + \frac{5\mu(t)}{250+4t}S = 27\mu(t)$$ and we want to realize the left hand side as the derivative of a product; that is, we want $$\mu'(t) = \frac{5\mu(t)}{250+4t}.$$ Separating variables we have $$\begin{align*} \frac{\mu'(t)}{\mu(t)} &= \frac{5}{250+4t}\\ \int\frac{d\mu}{\mu} &= \int \frac{5\,dt}{250+4t}\\ \ln|\mu| &= \frac{5}{4}\ln|250+4t| + C\\ \mu(t) &= A(250+4t)^{5/4} \end{align*}$$ Picking $A=1$, we can take $\mu(t) = (250+4t)^{5/4}$. (Another error in your computation).
That is, we have: $$(250+4t)^{5/4}S' + \frac{5(250+4t)^{5/4}}{250+4t}S = 27(250+4t)^{5/4}$$ or $$(250+4t)^{5/4}S' + 5(250+4t)^{1/4}S = 27(250+4t)^{5/4}$$ which can be written as $$\Bigl( (250+4t)^{5/4}S\Bigr)' = 27(250+4t)^{5/4}.$$
You might benefit from writing out the derivations very carefully (as I did above) rather than trying to rely on formulas (I assume that's how you tried to obtain your integrating factor $I$, which was mistakenly computed).
Can you take it from here? Careful with the integral on the right hand side.