$$\phi(x)=\int_{-\infty}^\infty\frac{1}{\pi}\frac{1}{(1+x^2)}{e^{itx}dx}$$
When you choose the contour, it depends on the sign of $t$, if $t>0$, you choose the upper half circle as the contour, because when you evaluate the integral on the upper half circle, you let $z=Re^{i\theta}$, where $\theta$ goes from $0$ to $\pi$ along the upper half circle counter-clockwisely. For the factor:
$$e^{itz}=e^{itR\cos\theta-tR\sin\theta},~~~~~t>0,~\sin\theta>0\tag{1}$$
as $R\to\infty$, the exponential factor in (1) will decay, hence the integral along the upper half circle vanishes. (We also use this feature when prove Jordan's lemma.)
In this case, only $z=i$ is inside the upper circle contour, hence you only evaluate the residue at $z=i$, so you get
$$\int_{-\infty}^\infty\frac{e^{itz}}{1+z^2}dz=2\pi i\cdot\frac{e^{-t}}{2\pi i}=e^{-t}$$
Similarly, if $t<0$, to guarantee the integral vanish on the half circle, we need to choose the lower half circle. Let $z=Re^{i\theta}$, where $\theta$ goes from $0$ to $-\pi$ along the lower half circle clockwisely, and
$$e^{itz}=e^{itR\cos\theta-tR\sin\theta},~~~~~t<0,~\sin\theta<0\tag{2}$$
Only the $z=-i$ is inside this contour, so you only compute the residue at $z=-i$. Hence, you get
$$\int_{-\infty}^\infty\frac{e^{itz}}{1+z^2}dz=\color{red}{-2\pi i}\cdot\frac{e^{t}}{-2\pi i}=e^{t}$$
Note the extra $-$ sign in the red-colored part, it is due to you go clockwisely when you do the integration along the lower half circle contour.
Best Answer
We could try a keyhole contour. Denote
$$f(z) = \frac{\ln(z)}{(az+b)(z+c)} \quad a,b,c> 0$$
and let $\ln(z) = \ln|z| +i\Theta \, $ where $0< \Theta\leq 2\pi$
Let $\epsilon, R >0$ such that $\epsilon <|\frac{b}{a}|<R$ and $\epsilon <|c|<R$. If we choose the branch cut over the $x$-axis, $0\leq x<\infty$ we have the following keyhole contour:
\begin{align*} \gamma_{1}(t) =& t, \, t\in(\epsilon, R)\\ \gamma_{2}(\theta) = &Re^{-i\theta}, \, \theta \in (-2\pi,0)\\ \gamma_{3}(t) =& -t, \, t\in(-R,-\epsilon)\\ \gamma_{2}(\theta) = &\epsilon e^{i\theta}, \, \theta \in (0,2\pi)\\ \end{align*}
If $\gamma = \gamma_{1}+...\gamma_{4}$, by the residue theorem: \begin{align*} \oint_{\gamma}f(z) dz = &\left(\oint_{\gamma_{1}}+...+\oint_{\gamma_{4}} \right) = \int_{\epsilon}^{R} \frac{\ln(t)}{(at+b)(t+c)}dt + \int_{R}^{\epsilon} \frac{\ln(e^{2\pi i}t)}{(at+b)(t+c)}dt +\oint_{\gamma_{2}}+\oint_{\gamma_{4}}\\ =& -2\pi i \int_{R}^{\epsilon} \frac{1}{(at+b)(t+c)}dt +\oint_{\gamma_{2}}+\oint_{\gamma_{4}}\\ =& 2\pi i \left(\operatorname{Res}\left(f,-\frac{b}{a}\right) + \operatorname{Res}(f,-c)\right)\\ =& 2\pi i \frac{\ln(b)-\ln(a)-i\pi -\ln(c) +i \pi}{ac-b}\\ =& \frac{2\pi i \ln\left(\frac{b}{ac}\right)}{ac-b} \end{align*}
Also note that
$$\left|\oint_{\gamma_{2}} f(z) dz \right|= \left|\int_{0}^{2\pi} \frac{iRe^{i\theta}\ln(Re^{i\theta})}{(aRe^{i\theta}+b)(Re^{i\theta}+c)}\right|\leq \frac{2\pi R(\ln(R)+2\pi)}{a(R-\left|\frac{b}{a}\right|)(R-|c|)}$$
$$\left|\oint_{\gamma_{4}} f(z) dz \right|= \left|\int_{0}^{2\pi} \frac{i\epsilon e^{i\theta}\ln(\epsilon e^{i\theta})}{(a\epsilon e^{-i\theta}+b)(\epsilon e^{i\theta}+c)}\right|\leq \frac{2\pi \epsilon (\ln(\epsilon)+2\pi)}{a(\epsilon-\left|\frac{b}{a}\right|)(\epsilon-|c|)}$$
So, when $\epsilon \to 0+$ and $R \to \infty$
$$ \oint_{\gamma_{2}} f(z) dz + \oint_{\gamma_{4}}f(z) dz \to 0 $$
Then
$$ \int_{0}^{\infty} \frac{1}{(at+b)(t+c)}dt = \frac{ \ln\left(\frac{ac}{b}\right)}{ac-b}$$
Now differentiating both sides $k-1$ times with respect to $a$:
$$\int_{0}^{\infty} \frac{x^{k-1}(-1)^{k-1}(k-1)!}{(xa+b)^{k}(x+c)}dx = \frac{(-1)^{k-1}(k-1)!c^{k-1}\ln\left(\frac{ac}{b}\right)}{(ac-b)^{k}}- \frac{1}{(ac-b)^{k}} \sum_{j=1}^{k-1} \binom{k-1}{j} \frac{(k-1-j)!(ac-b)^{j}c^{k-1-j}(j-1)!}{a^j} $$
Hence
$$\int_{0}^{\infty} \frac{x^{k-1}}{(xa+b)^{k}(x+c)}dx = \frac{c^{k-1}\ln\left(\frac{ac}{b}\right)}{(ac-b)^{k}}- \frac{1}{(ac-b)^{k}} \sum_{j=1}^{k-1} \frac{(ac-b)^{j}c^{k-1-j}}{ja^j} $$
if we put $a=1$
$$\boxed{\int_{0}^{\infty} \frac{x^{k-1}}{(x+b)^{k}(x+c)}dx = \frac{c^{k-1}\ln\left(\frac{c}{b}\right)}{(c-b)^{k}}- \frac{1}{(c-b)^{k}} \sum_{j=1}^{k-1} \frac{(c-b)^{j}c^{k-1-j}}{j} \quad b,c>0, k\in \mathbb{N}} $$
You can continue increasing the powers in the denominator of the left hand side differentiating with respect to $b$ both sides of the equation.
For example, differentiating one more time both sides with respect to $b$
$$\boxed{\int_{0}^{\infty} \frac{x^{k-1}}{(x+b)^{k+1}(x+c)}dx = -\frac{c^{k-1}\ln\left(\frac{c}{b}\right)}{(c-b)^{k+1}}+\frac{c^{k-1}}{kb(c-b)^{k}}+ \frac{1}{k(c-b)^{k+1}} \sum_{j=1}^{k-1} \frac{(k-j)(c-b)^{j}c^{k-1-j}}{j} \quad b,c>0, k\in \mathbb{N}} $$