Prove that the linear functional $$ A(f) = \int_2^4 \left( x^3f(x) + \dfrac{if(x+1)}{x^2} \right)dx \quad \forall f \in L_2(\mathbb{R}) $$ is continuous and calculate its norm.
To solve this problem I computed the norm of $A$ directly. By doing so, we would prove that it is bounded, and since it is linear, that would imply its continuity. The way I tried to do this is by using the Riesz theorem. If I find a $g\in L_2(\mathbb{R})$ such that $A(f) = \langle g,f \rangle$, then $\|A\| = \|g\|$.
$$ \begin{align} A(f) &= \int_2^4 x^3 f(x)\ dx + \int_2^4 \dfrac{if(x+1)}{x^2}\ dx = \int_2^4 x^3 f(x)\ dx + \int_3^5 \dfrac{if(x)}{(x-1)^2}\ dx \\ &=\int_2^3 x^3 f(x)\ dx + \int_3^4 \left( x^3 + \dfrac{i}{(x-1)^2} \right) f(x)\ dx + \int_4^5 \dfrac{if(x)}{(x-1)^2}\ dx. \end{align} $$
Defining a function $g$ as $$ g(x) = \left\{\begin{array}{cc} x^3 & \text{if } x \in [2,3] \\ x^3 - \dfrac{i}{(x-1)^2} & \text{if } x \in (3,4] \\ - \dfrac{i}{(x-1)^2} & \text{if } x \in (4,5] \\ 0 & \text{in any other case} \end{array}\right. $$ it is clear that $A(f) = \langle g,f \rangle$, so $$ \begin{align} \|A\|^2 &= \|g\|^2 = \int_{\mathbb{R}} |g(x)|^2 dx = \int_2^3 x^6\ dx + \int_3^4 \left(x^6 + \dfrac{1}{(x-1)^4}\right)dx + \int_4^5 \dfrac{1}{(x-1)^4}\ dx. \end{align} $$
The norm $\|A\|$ is finite, so the functional is bounded and, therefore, continuous.
Is this a correct approach? Have I make any mistake?