A standard way forward to evaluate an integral such as $\displaystyle \int_0^\infty \frac{\log(x)}{(x+a)(x+b)}\,dx$ using contour integration is to evaluate the contour integral $\displaystyle \oint_{C}\frac{\log^2(z)}{(z+a)(z+b)}\,dz$ where $C$ is the classical keyhole contour.
Proceeding accordingly we cut the plane with a branch cut extending from $0$ to the point at infinity along the positive real axis. Then, we have
$$\begin{align}
\oint_{C} \frac{\log^2(z)}{(z+a)(z+b)}\,dz&=\int_\varepsilon^R \frac{\log^2(x)}{(x+a)(x+b)}\,dx\\\\
& +\int_0^{2\pi}\frac{\log^2(Re^{i\phi})}{(Re^{i\phi}+a)(Re^{i\phi}+b)}\,iRe^{i\phi}\,d\phi\\\\
&+\int_R^\varepsilon \frac{(\log(x)+i2\pi)^2}{(x+a)(x+b)}\,dx\\\\
&+\int_{2\pi}^0 \frac{\log^2(\varepsilon e^{i\phi})}{(\varepsilon e^{i\phi}+a)(\varepsilon e^{i\phi}+b)}\,i\varepsilon e^{i\phi}\,d\phi\tag1
\end{align}$$
As $R\to \infty$ and $\varepsilon\to 0$, the second and fourth integrals on the right-hand side of $(1)$ vanish and we find that
$$\begin{align}\lim_{R\to\infty\\\varepsilon\to0}\oint_{C} \frac{\log^2(z)}{(z+a)(z+b)}\,dz&=-i4\pi \int_0^\infty \frac{\log(x)}{(x+a)(x+b)}\,dx\\\\
&+4\pi^2\int_0^\infty \frac{1}{(x+a)(x+b)}\,dx\tag2
\end{align}$$
And from the residue theorem, we have for $R>\max(a,b)$
$$\begin{align}
\oint_{C} \frac{\log^2(z)}{(z+a)(z+b)}\,dz&=2\pi i \left(\frac{(\log(a)+i\pi)^2}{b-a}+\frac{(\log(b)+i\pi)^2}{a-b}\right)\\\\
&=2\pi i\left(\frac{\log^2(a)-\log^2(b)}{b-a}\right)-4\pi ^2 \frac{\log(a/b)}{b-a} \tag3
\end{align}$$
Now, finish by equating the real and imaginary parts of $(2)$ and $(3)$.
Can you finish now?
$$J = \int_0^\infty \frac{ \log x \, dx} {(1+x)^3}.$$
Consider $$\oint_C \frac{(\log z)^2 \, dz}{(1+z)^3}$$ around a suitable keyhole contour $C$ that starts at $\epsilon$ goes to $R$, a large (almost) circle of radius $R$, back (below the branch cut) to $\epsilon$ and then clockwise around the origin.
There is a third order pole inside at $z_0 = -1$. The residue there is $$\text{Residue}_{z=-1} \left[\frac{ (\log z)^2}{(1+z)^3}\right] = 1-i\pi.$$
$$\begin{aligned}
\oint_C \frac{(\log z)^2 \, dz}{(1+z)^3} &=
\int_\epsilon^\infty \frac{(\log x)^2 \, dx}{(1+x)^3} -\int_\epsilon^\infty \frac{(\log x+2i\pi)^2 \, dx}{(1+x)^3}+\int_0^{2\pi} \frac{(\log (Re^{i\theta}))^2 \, Rie^{i\theta} }{(1+Re^{i\theta})^3}\, d\theta
-~\int_0^{2\pi} \frac{(\log (\epsilon e^{i\theta}))^2 \, \epsilon i \, e^{i\theta} }{(1+\epsilon e^{i\theta})^3}\, d\theta
\end{aligned} $$
Let $R\to\infty$ and $\epsilon\to 0$. The integrals along the "circles" go to zero.
Also, $$\displaystyle \int_0^\infty \frac{dx}{(1+x)^3}=\int_1^\infty \frac{dp}{p^3} = \left. -\frac{p^{-2}}{2} \right|_1^\infty = \frac{1}{2}.$$
So, we have
$$-4i\pi J + 4\pi^2 \left( \frac{1}{2}\right) = 2\pi i (1-i\pi).$$
$$J=-\frac{1}{2}$$
Best Answer
You're missing the outer circle to close the loop, which diverges. Furthermore, the rays around the positive real line will cancel out, so the given integral does not simply equal the integrals you wrote.
A better integral to use would be a semicircle contour avoiding the origin. Note also that the integrand is symmetric:
$$\int_0^\infty\frac{\sin(x)}{x^3+x}~\mathrm dx=\frac12\int_{-\infty}^\infty\frac{\sin(x)}{x^3+x}~\mathrm dx$$
The inner (clockwise semi-) circle will converge to $-\pi i$ while the outer circle will vanish, leaving us with
$$-\pi i+\int_{-\infty}^\infty\frac{e^{iz}}{z^3+z}~\mathrm dz=\oint_C\frac{e^{iz}}{z^3+z}~\mathrm dz=2\pi i\underset{z=i}{\operatorname{Res}}\frac{e^{iz}}{z^3+z}$$
and you should be able to take it from here.