<s>[INST] Partial fractions to integrate$\int \frac{4x^2 -20}{(2x+5)^3}dx$ $$\int \frac{4x^2 -20}{(2x+5)^3}dx$$
I can't use the coverup method that I learned since making anything zero in this makes everything zero. I would probably just use random test points because I don't have any other tricks memorized. Is there some specific trick I need to use on this so it is possible to compute on a timed test?
 [/INST] $$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac A{(2x + 5)} + \dfrac B{(2x + 5)^2} + \dfrac C{(2x + 5)^3}\, dx$$
Partial fractions can be thought of as the reverse of "finding the common denominator".
That is, we can equate the numerator of the original integrand with the expansion of each of the numerators of the desired fractions, times the factor we'd need to find the common denominator, if we were adding the fractions in the desired integrand.
Doing this gives us $$A(2x + 5)^2 + B(2x + 5) + C = 4x^2 - 20\tag{1}$$
Solve for C first: to do this we can put $x = -\frac 52$ to zero out all but $$C = 4(-5/2)^2 - 20 = 4\cdot \frac {25}{4} - 20 = 25 - 20 = 5$$ 
Then expand the left-hand side, substitute $C = 5$, and match up coefficients, (you can easily find $A$ that way, since it will be the only term on the left hand side that is a coefficient of a term with $x^2$. Substitute, solve for $B$ knowing $A$. In other situations, you could use two arbitrary values for $x$ to obtain two equations in two unknowns. 
Expanding $(1)$: $$A(4x^2 + 20 x + 25) + B(2x + 5) + 5 = 4x^2 - 20 \tag{2}$$ 
$$\implies  4Ax^2 = 4 \implies A = 1$$
Now, we have $A = 1, C = 5$, and can substitute this into our equation $(2)$:
$$4Ax^2 + 20 Ax + 25A + 2Bx + 5B + 5 = 4x^2 - 20$$  $$ \iff (20 A + 2B)x = 0\cdot x $$ $$\iff 20 A + 2B = 0 $$ $$\iff 20\cdot 1 + 2B = 0 \iff B = -10 $$
$$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac 1{(2x + 5)} + \dfrac {-10}{(2x + 5)^2} + \dfrac 5{(2x + 5)^3}\, dx$$
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