We are seeking to evaluate this integral $(1)$
$$\int_{0}^{\infty}\left((1+x)^{-n}-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x},n\ge1\tag1$$
$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$
$$\int_{0}^{\infty}\left(1+nx+\frac{n(n-1)}{2!}x^2+\cdots-\frac{\sin x}{x}\right)\frac{\mathrm dx}{x}$$
$$\int_{0}^{\infty}\left(x^{-1}+n+\frac{n(n-1)}{2!}x+\cdots-\frac{\sin x}{x^2}\right)\mathrm dx$$
we run into this integral which is not possible $\int_{0}^{\infty}\frac{1}{x}\mathrm dx $
First, integrate by parts on the $\sin(x)/x$ term $$ I = \int_{0}^{\infty}\left[(1+x)^{-n}-\frac{\sin x}{x}\right]\frac{dx}{x} = \int_{0}^{\infty}\left[(1+x)^{-n}-\cos(x)\right]\frac{dx}{x} - 1 $$ Next, introduce the term $e^{-x}/x$ into the integral \begin{multline} I = \int_{0}^{\infty}\left[(1+x)^{-n} - e^{-x} + e^{-x}+\cos(x)\right]\frac{dx}{x} - 1 \\= \int_{0}^{\infty}\left[(1+x)^{-n} - e^{-x}\right]\frac{dx}{x} + \int_0^\infty \left[e^{-x}-\cos(x)\right]\frac{dx}{x} - 1 \\= \int_0^\infty \frac{e^{-x}-\cos(x)}{x}dx - \psi(n)-1 \end{multline}
Where the last equality comes from the integral representation of the digamma function $\psi$ found here. Now, as it happens, $\int_0^\infty [e^{-x}-\cos(x)]dx/x = 0$. This can be shown in a few ways, such as showing the Laplace transform of the integrand is $\ln[\sqrt{1+s^2}/(1+s)]$. At any rate, this gives $$ \int_{0}^{\infty}\left[(1+x)^{-n}-\frac{\sin x}{x}\right]\frac{dx}{x} = -\psi(n) -1 . $$
