If _1^2 dx(x^2-2x+4)^3/2 = kk+5 then k is equal to:

Question

If $\displaystyle\int_1^2 \dfrac{dx}{(x^2-2x+4)^{3/2}} = \dfrac{k}{k+5}$ then $k$ is equal to:

Accepted Answer

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

$$ \Rightarrow $$ $\,\,\,$

Solutions · 1