Question: Is the following question poorly written? 

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> [![enter image description here](https://i.stack.imgur.com/3uOIp.png)](https://i.stack.imgur.com/3uOIp.png)
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I suppose it is a translation since it comes from a Japanese institution. I am referring to the assumption "... has a solution". Given this information I would assume that the discriminant is equal to $0$ since the equation has "a solution" but the answer involves writing it with a "greater than or equal to" symbol. 

What are your opinions on this? Am I wrong or should the translation specify that it has one or two solutions?

Answer: 
Need to split the interval in $[-1,0];[0,1]$

Method $\#1:$

Use [Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?](https://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function)

For $x>0,$

$\arctan\dfrac1x=$arccot$(x)=\dfrac\pi2-\arctan(x)$

For $x<0,$

$\arctan\dfrac1x=-\arctan\left(-\dfrac1x\right)=-$arccot$(-x)=-(\pi-$arccot$x)=-\pi+\dfrac\pi2-\arctan(x)$

Method$\#2:$

Alternatively using [Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$](https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan)

$\arctan x+\arctan\dfrac1x=$sign$(x)\cdot\dfrac\pi2$
Here we will address your integral:
\begin{equation}
 \int\_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx
\end{equation}
Now:
\begin{equation}
 \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right] = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot -\frac{1}{x^2} = -\frac{1}{1 + x^2}
\end{equation}
Thus your integral becomes:
\begin{align}
 \int\_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx &= \int\_{-1}^{1} -\frac{1}{1 + x^2}\:dx = -2\int\_0^1 \frac{1}{1 + x^2}\:dx \nonumber \\
&= -2\left[\arctan(x) \right]\_0^1 = -2 \cdot \frac{\pi}{4} = -\frac{\pi}{2}
\end{align}