I'm trying to find the length of the positive half of an ellipse that is centered at $(.75,0)$ with a horizontal axis of $1.5$ and a vertical axis of $.5$. The formula for this arc would be:
$$\frac{\left(x-.75\right)^2}{1.5^2}+\frac{y^2}{.5^2}=1$$ $$.25-.25\left[\frac{(x-.75)^2}{2.15}\right]=y^2$$ $$.25-.5375\left(x^2-1.5x-.5625\right)=y^2$$ $$\sqrt{-\frac{43}{80}x^2+\frac{129}{160}x+\frac{707}{1280}}=y$$ $$\sqrt{\frac{1051}{1280}\left(x+1\right)^2}=y$$
The length of an arc is $$L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx.$$ In our case, $a=0$, $b=1.5$, and $$\frac{dy}{dx}=\frac1{2\sqrt{\frac{1051}{1280}\left(x+1\right)^2}}\cdot2\left[\frac{1051}{1280}\left(x+1\right)\right]=\frac{\frac{1051}{1280}\left(x+1\right)}{2\sqrt{\frac{1051}{1280}\left(x+1\right)^2}}.$$
Plugging that in,
$$L=\int_0^{1.5}\sqrt{1+\left(\frac{\frac{1051}{1280}\left(x+1\right)}{2\sqrt{\frac{1051}{1280}\left(x+1\right)^2}}\right)^2}dx.$$
This is where I'm stuck. That's an awfully-looking difficult integral to deal with, and I don't even know where to begin for integration by parts. Any advice?