I know that length of the curve is either of: $$ s = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx = \int_c^d \sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy$$ Now the curve is $y=x^2$, $-1\leq x\leq2$, when I use the first formula, I get $$s=\frac{1}{6}\left[(1+4x)^{3/2}\right]^{2}_{-1}$$ When I put the lower limit, it gives imaginary number, how to handle this and calculate the length of the curve from $x=-1$ to $x=2$.
The second question is: I have to calculate the length of the curve $y^2+2y=2x+1$, from the point $(-1,-1)$ to the point on the curve $(7,3).$ How to calculate the length in these case, I know how to calculate in simple cases but can't handle these!
In your first one, you did not square the derivative and integrated the wrong function, as is pointed out by Randall.
In the second one, consider $x = x(y) = (y^2+2y-1)/2 = y^2/2+y-1/2$, therefore $x'(y) = y+1$ and $(x'(y))^2=(y+1)^2$. Now use the second form you yourself provided...