Solve $\int\frac{8x+9}{(2x+1)^3}\,dx$.

Question

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 $\dfrac{-16x+13}{4(2x+1)^2} +C$

Answer 1

You can simply use $u=2x+1$ and you will get a sum of simple powers that you can integrate with ease.

Answer 2

Hint $$\frac{8x+9}{(2x+1)^3} = \frac{4(2x+1)}{(2x+1)^3}+\frac5{(2x+1)^3} = \frac{4}{(2x+1)^2}+\frac5{(2x+1)^3}$$

Answer 3

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.

Answer 4

write the numerator of the integrand as $$8x+4+5=4(2x+1)+5$$ then we have $$4\int\frac{dx}{(2x+1)^2}dx+5\int\frac{dx}{(2x+1)^3}dx$$ now you can set $$u=2x+1$$ and proceed.