Consider the following integral :
$$I(x;a)= \int \frac{\sin^2(x+1)}{(x+1)^a}\,dx$$
Now let the $x$ runs from $0$ to $\infty$
i.e. let
$$I_p(a)= \int_0^\infty \frac{\sin^2(x+1)}{(x+1)^a}\,dx$$
Now we can see that $I_p(a)$ converges for all $\text{R}(a)>1$.
Also, for $a=1$ , $I_p(a)$ diverges .
Question 1 : If possible , can $I_p(a)$ be represented in following way :
$$I_p(a) = f(a) + \frac{C}{(a-1)}$$
Here , $f(a)$ is convergent for all $a\geqslant1$ and $C>0$
The factor $\frac{C}{(a-1)}$ explains the divergence of $I_p(a)$ at $a=1$
Question 2 ( important ):
Analogous to Question 1 : Only replace $\sin^2(x+1)$ by $\sin^2(g(x))$ in $I_p(a)$
where, $g(x)$ is an increasing function in $[1,\infty)$
For the first part, use the double angle formula to have $$\sin^2(x+1)=\frac 12 \left(1-\cos(2(x+1))\right)$$ Then use $\cos(t)=\frac {e^{it}+e^{-it}}2$ to face the exponential integral function.
You should end with something like $$I(a)=\int_0^\infty \frac{ \sin ^2(x+1)} {(x+1)^{a}}\,dx=\frac{1}{2 (a-1)}-\frac{1}{4} (E_a(2 i)+E_a(-2 i))$$ provided $\Re(a)>1$.