<s>[INST] Solve $\int\frac{8x+9}{(2x+1)^3}\,dx$. Do I split  $\displaystyle\int\frac{8x+9}{(2x+1)^3}\,dx$  into partial fractions? Or do I use $(2x+1)^{-3}$ by itself? Not sure what to do. Please advice. 
The answer given is $\dfrac{-16x+13}{4(2x+1)^2} +C$
 [/INST] Partial fractions is overkill for this problem. 
Substitute $u = 2x+1$ to get $\displaystyle\int\dfrac{8x+9}{(2x+1)^3}\,dx = \int\dfrac{4u+5}{u^3}\dfrac{du}{2} = \dfrac{1}{2}\int\left(\dfrac{4}{u^2}+\dfrac{5}{u^3}\right)\,du$. 
This is easy to handle using the power rule.
Remark: Simplifying $\dfrac{4u+5}{u^3} = \dfrac{4u}{u^3}+\dfrac{5}{u^3} = \dfrac{4}{u^2}+\dfrac{5}{u^3}$ is essentially equivalant to what you would have gotten by using partial fractions.
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