# the solid lies between planes perpendicular to the x-axis at x=-1 and x=1. the cross-sections perpendicular to the x-axis are a. circles whose diameters stretch from the curve y=-1/(sqrt(1+x^2)) to the curve y=1/(1+x^2) b. vertical squares whose base edges run along the same ends

the solid lies between planes perpendicular to the x-axis at x=-1 and x=1. the cross-sections perpendicular to the x-axisare
a. circles whose diameters stretch from the curve $y=-\frac{1}{\sqrt{1+{x}^{2}}}$ to the curve $y=\frac{1}{\sqrt{1+{x}^{2}}}$
b. vertical squares whose base edges run along the same ends
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suponeriq
Have you had the derivatives of inverse trigonometric functions yet?
$\frac{d}{dx}\left[{\mathrm{tan}}^{-1}\left(x\right)\right]=\frac{1}{1+{x}^{2}}$
If so you can use the definition of the anti-derivative to deduce
$\frac{d}{dx}\left[{\mathrm{tan}}^{-1}\left(x\right)\right]=\frac{1}{1+{x}^{2}}⇒\int \frac{1}{1+{x}^{2}}dx={\mathrm{tan}}^{-1}\left(x\right)$