Finding center of mass of a thin plate of constant density in specific region

Question

The region is between $ y= \frac{1}{1+x^2}$ and $ y= -\frac{1}{1+x^2} , x=0 , x=1$

What I currently have is this:

The answer is $ \bar {x} = \frac{\ln 4}{\pi} $ but I cannot seem to get that. Was I supposed to have used $\tan^{-1}x$ ? I'm lost.

Answer 1

Notice, the area bounded by the given curves: $y=\frac{1}{1+x^2}$ & $y=-\frac{1}{1+x^2}$ between $x=0$ & $x=1$, is symmetrical about the x-axis hence the center of mass will lie on the x-axis.

Now, the x-coordinate of the center of mass of the bounded region is given as

$$\bar{x}=\frac{\int x\ dA}{\int dA}=\frac{\int x(ydx)}{\int (ydx)}$$ Setting the value of $y$ & applying proper limits, $$=\frac{2\int_{0}^{1}x(ydx)}{2\int_{0}^{1}(ydx)}$$$$=\frac{\int_{0}^{1}x\frac{1}{1+x^2}\ dx}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$ $$=\frac{\int_{0}^{1}\frac{1}{2}\frac{2x}{1+x^2}\ dx}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$$$=\frac{\frac{1}{2}\int_{0}^{1}\frac{d(x^2)}{1+x^2}}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$ $$=\frac{1}{2}\frac{[\ln(1+x^2)]_{0}^{1}}{[\tan^{-1}(x)]_{0}^{1}}$$ $$=\frac{1}{2}\frac{[\ln(2)-0]}{[\tan^{-1}(1)-0]}=\frac{1}{2}\frac{\ln 2}{\frac{\pi}{4}}=\frac{2\ln 2}{\pi}$$$$\color{red}{\bar x=\frac{\ln 4}{\pi}}$$