Answer
$$\int^{1}_0\frac{36dx}{(2x+1)^3}=8$$
Work Step by Step
$$A=\int^{1}_0\frac{36dx}{(2x+1)^3}$$
We set $a=2x+1$, which means $$da=2dx$$ $$dx=\frac{1}{2}da$$
For $x=0$, $a=1$ and for $x=1$, $a=3$
Therefore,
$$A=\frac{1}{2}\int^{3}_1\frac{36}{a^3}da=18\int^{3}_1a^{-3}da$$ $$A=\Big(\frac{18a^{-2}}{-2}\Big)\Big]^{3}_1=\Big(-\frac{9}{a^2}\Big)\Big]^{3}_1$$ $$A=-1-(-9)=-1+9$$ $$A=8$$
