If f(x) is given, what would be the distribution of Y = 2X + 1? 
If the random variable X has a continuous distribution with the
  density $ f(x) = \frac{1}{x^2}\Bbb {1}_{(1, ∞)}(x)$, can you find the
  distribution of $Y = 2X+1$?

My attempt:
$CDF(Y)$
$\Rightarrow P(Y\le y)$
$\Rightarrow P(2X + 1 \le y)$
$\Rightarrow P(X \le \frac{y-1}{2})$
$\Rightarrow \int_{1}^{(y-1)/2} f(x) dx$
$\Rightarrow \int_{1}^{(y-1)/2} \frac{1}{x^2} \Bbb 1_{(1, \infty)}(x) dx$
$\Rightarrow \int_{1}^{(y-1)/2} \frac{1}{x^2} dx$
$\Rightarrow \frac{1}{-2+1}[x^{-2+1}]_1^{(y-1)/2}$
$\Rightarrow 1 - \frac{2}{y-1}$ Ans.

Is it a correct approach?

Since the transformation function is monotonic, we can find the CDF by using PDF transformation and integrating the transformed PDF.
PDF Transformation: 
$$ f_Y(y) = f_X(g^{-1}(y)) \Bigg|\frac{dg^{-1}}{dy} \Bigg|$$
For this situation, $g^{-1}(y) = \frac{y-1}{2}$, and by substitution:
$$f_X(g^{-1}(y)) = \frac{1}{((y-1)/2)^2} = \frac{4}{(y-1)^2}$$ 
The absolute value of the derivative of $g^{-1}(y)$ with respect to $y$ is easy:
\begin{align}
\Bigg|\frac{dg^{-1}}{dy} \Bigg| = \frac{1}{2}
\end{align}
Plug components into the PDF Transformation formula above to get a transformed PDF of :
\begin{align}
f_Y(y) = \frac{2}{(y-1)^2} \quad \text{for} \quad 3 \leq y \lt \infty
\end{align}
Remembering to use transformed lower and upper bounds, we integrate to get the CDF. Line 3 employs u-substitution to simplify the integration:
\begin{align} 
F_Y(y) &=\int_{3}^{y} f_Y(t) \, dt\\  
&=\int_{3}^{y} \frac{2}{(t-1)^2} \, dt\\  
&=\int_{2}^{y-1} \frac{2}{u^2} \, du\\
&= -\frac{2}{u} \; \Bigg|_2^{y-1}\\
&= -\frac{2}{y-1} - (-\frac{2}{2})\\ 
&= 1 - \frac{2}{y-1} \quad \text{for} \quad 3 \leq y \lt \infty
\end{align}
Finally, we'll check the lower and upper bounds just to make sure we've fulfilled CDF validity requirements:
$$F_Y(3) = 1 - \frac{2}{2} = 0$$
$$F_Y(\infty) = 1 - \frac{2}{\infty} = 1$$
