Since x^2-2x+4 = (x-1)^2+3, we substitute
x-1 = sqrt{3} tan theta, qquad dx = sqrt{3} sec^2 theta
Again, @ x=1 we have theta=0 and @x=2, we have
theta = tan^{-1}(1/sqrt{3})=pi/6. Thus, the integral becomes
int_1^2 dx/(x^2-2x+4)^{3/2} = int_0^{pi/6} {sqrt{3} sec^2 theta d theta}/{3sqrt{3}sec^3 theta} = 1/3int_0^{pi/6}cos theta d theta = 1/3 sin theta|_0^{pi/6}=1/6
Hence
k/{k+5} = 1/6 implies k=1