$\int_{1}^{2}\frac{1}{(3-5t)^2}dt$ Let $g(t)=(3-5t)^2$, $f(x)=\frac{1}{x^2}$, $g(1)=-2$, $g(2)=-7$.     
$$
\begin{align}
& \phantom{={}}\int_1^2 \frac{1}{(3-5t)^2} \, dt \\[6pt]
& =-\frac{1}{5}\int_1^2 \frac{1}{f(g(t))}g'(t) \, dt \\[6pt]
& =-\frac{1}{5}\int_{-2}^{-7}\frac{1}{x^2} \, dx \\[6pt]
& =-\frac{1}{5}\left[-\frac{1}{x}\right]_{-2}^{-7}
\end{align}
$$
I can't follow the whole process. Why we have to multiply $g'(t)$ and why $\dfrac{1}{f(g(t))}$?
Also in the third line, why does $g'(t)$ disappear? $-\frac{1}{x}$ is the correct one?
 A: Why not directly?
$$(3-5t)'=-5\implies \int\limits_1^2\frac{dt}{(3-5t)^2}=-\frac15\int\limits_1^2\frac{(-5dt)}{(3-5t)^2}=$$
$$=\left.\frac15\frac1{3-5t}\right|_1^2=\frac15\left(-\frac17+\frac12\right)=\frac1{14}$$
The above is based on
$$\int\frac{f'}{f^2}=-\frac1f+C$$
A: The procedure outlined makes sense only if $g(t) = 3 - 5t$, and not, as you've written, $g(t) = (3-5t)^2$. 
So given $u = g(t) = 3 - 5t,\;$ we know that $\;g(1) = 3 - 5(1) = -2,\;$ and $\;g(2) = 3-5(2) = -7$. 
Then given $f(x) = \dfrac 1{x^2}$, it follows that $\;f(g(t)) = \dfrac 1{(g(t)^2)} = \dfrac 1{(3 - 5t)^2}$. 
So the substitution being made is $g(t) = \color{red}{\bf u = 3 - 5t}$. 
Then $g'(t)  = du = -5 dt \iff \color{blue}{\bf -\frac 15 du = dt.}$
$$\int_1^2 \dfrac{1}{(\color{red}{\bf 3 - 5t})^2} \color{blue}{\bf\,dt} = \int_{-2}^{-7} \dfrac{1}{\color{red}{\bf u}^2}\,\cdot\left(\color{blue}{\bf -\frac 15 du}\right)= -\frac 15\int_{-2}^{-7} \dfrac{1}{u^2}\,du $$
A: First of all, your $g(t)$ should be $3-5t,$ not $(3-5t)^2$. Also, it should be $f(g(t)),$ not $\frac1{f(g(t))}.$
The idea, here, is to make a substitution so that the integrand is "nicer." The idea was to let $x=g(t),$ since then $$\frac1{x^2}=\frac1{(3-5t)^2}.$$ Now, since $x=g(t),$ then $\frac{dx}{dt}=g'(t),$ and so $$dx=g'(t)dt.$$ Since $g'(t)=-5,$ and since $x$ ranges from $-2$ to $-7$ as $t$ ranges from $1$ to $2$, then $$\begin{align}\int_0^2\frac1{(3-5t)^2}\,dt &= \int_0^2f\bigl(g(t)\bigr)\,dt \\ &= \int_0^2f\bigl(g(t)\bigr)\cdot\frac{g'(t)}{-5}\,dt\\ &=-\frac15\int_0^2f\bigl(g(t)\bigr)\cdot g'(t)\,dt\\ &= -\frac15\int_{-2}^{-7}f(x)\,dx\\ &= -\frac15\int_{-2}^{-7}\frac1{x^2}\,dx\end{align}$$ by substitution and our work above.
