$$\int_1^2 \frac{dx}{3-5x^2}$$
Step 1: let u = 3-5x and -1/5 du = dx.
when x=2, u = -7 and when x=1, u = -2.
So now the integral becomes,
$$ -{1}/{5} \int_{-2}^{-7} \frac{1}{u^2} du$$
and the integral of,
$$ \int\frac{1}{u^2} = -\frac{1}{u}$$
So,
$$-\frac{1}{5} *-\frac{1}{u} = (\frac{1}{5}) * \frac{1}{u} $$
then plugging in the upper and lower limits for u,
$$\frac{1}{5} * (\frac{7}{14} - \frac{2}{14}) = \frac{1}{14} $$
I followed the solution from my textbook and understand what they did, however, the main problem I have is when they rewrote the integral in terms of u as,
$$ -{1}/{5} \int_{-2}^{-7} \frac{1}{u^2} du$$
Specifically, the $$ \frac{1}{u^2} $$
is what bothers me most because
$$ u^2 = (3-5x)^2...$$ NOT $$3-5x^2 $$
So can someone please explain to me why the textbook rewrote the integral as $$1/u^2$$??