Improper Integral of $\int\frac{dx}{(2x-1)^3}$ Improper Integral of 

$$\int_{-\infty}^0\frac{dx}{(2x-1)^3}$$ 

from Anton Calculus 8th Edition, page 576, question 9. Answer is $-\frac{1}{4}$ but I'm finding $-1$
The integral, substituting $u= (2x-1)$:
$$\frac{1}{2}\cdot\frac{-2}{(2x-1)^2}+C$$
Definite solution to use with limit:
$$ \frac{1}{2}\cdot\frac{-2}{(2x-1)^2)}\Big|^0_a = \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right)$$
Then solving limit:
$$\lim_{x \to {-\infty}} \frac{1}{2}\cdot\left(-2+\frac{2}{(2a-1)^2}\right) = \frac{1}{2}\cdot(-2+0) = -1$$
Any leads?
Thank you.

You made correct substitution
$$u= 2x-1\iff du=2dx$$
$$\int\frac{dx}{(2x-1)^3}dx=\frac12\int\frac{du}{u^3}dx=-\frac{1}{4u^2}+C$$

$$\int\frac{dx}{(2x-1)^3}=-\frac{1}{4(2x-1)^2}+C$$

I hope you can take it from here!
