Question

The value of 'k' is=?

Answer 1

1

Explanation:

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`