Opposite Operators Cancel Each Other, Right?

$\begin{aligned} I & = \int_{-1}^1 \frac d{dx} \left( \tan^{-1} \frac 1x \right) dx & \small {\color{#3D99F6}\text{see Note}} \\ & = \int_{-1}^1 -\frac 1{x^2+1} dx \\ & = \tan^{-1} x \ \bigg|_1^{-1} \\ & = - \frac \pi 4 - \frac \pi 4 \\ & = - \frac \pi 2 \approx \boxed{-1.5708} \end{aligned}$

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Note:

Let $\theta = \tan^{-1} \dfrac 1x$ .

$\begin{aligned} \implies \tan \theta & = \frac 1x \\ \sec^2 \theta \frac {d \theta}{dx} & = - \frac 1{x^2} \\ \frac {d \theta}{dx} & = - \frac 1{x^2} \times \frac 1{\sec^2 \theta} \\ \frac d{dx} \left( \tan^{-1} \frac 1x \right) & = - \frac 1{x^2} \times \frac 1{1 + \tan^2 \theta} \\ & = - \frac 1{x^2} \times \frac 1{1 + \frac 1{x^2}} \\ & = - \frac 1{x^2} \times \frac {x^2}{x^2+1} \\ & = - \frac 1{x^2+1} \end{aligned}$