Where did du/8x go for the integral? [closed]
Ok, I have a very simple question. I am trying to do the following integral by substitution. $$ \int \frac{x}{{ \left(4x^2 + 1\right)}^5 }dx $$
Putting it into an online solver, they substitute $u= 4x^2 + 1$ so they get $dx = \frac{1}{8x} du$. In the next step, they write $$ \frac{1}{8} \int \frac{1}{u^5} du $$ So my question is what happens to $x$ (as is $ \frac{1}{8x}$) as in in this step?
$$\int \frac{x}{u^5}*\frac{1}{8x}du$$ x in the numerator cancels out with x in the denominator