Given, I = $$\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} $$

= $$\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} $$

Let x $$-$$ 1 = $$\sqrt 3 $$ tan$$\theta $$

$$ \Rightarrow $$$$\,\,\,$$ dx = $$\sqrt 3 $$ sec²$$\theta $$ d$$\theta $$

When x = 1, then $$\theta $$ = 0

and when x = 2, $$\theta $$ = $${\pi \over 6}$$

I = $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}} $$

= $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}} $$

= $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}} $$

= $${1 \over 3}$$ $$\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}} $$

= $${1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta } $$

= $${1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}$$

= $${1 \over 3} \times {1 \over 2}$$

= $${1 \over 6}$$

$$\therefore\,\,\,$$ According to the question,

$${k \over {k + 5}}$$ = $${1 \over 6}$$

$$ \Rightarrow $$$$\,\,\,$$ 6k = k + 5

$$ \Rightarrow $$$$\,\,\,$$ k = 1