<s>[INST] 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.
 [/INST] Rather than multiplying by $-2,$ you should be dividing, since $$\int u^{n}\,du=\frac1{n+1}u^{n+1}$$ for all integers $n\ne-1.$
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