Question

${\int }_1^2\frac{dx}{{(x^2-2x+4)}^{\frac{3}{2}}}=\frac{k}{k+5}$, then $k$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

${\int }_1^2\frac{dx}{(x^2-2x+1{)}^{2/3}}={\int }_1^2\frac{dx}{\left[\right(x-1{)}^2+3{]}^{3/2}}$

Substitute $x-1=\sqrt{3}\tan \theta$ when $x=2$,

when $x=1,\theta =0$ $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$

$dx=\sqrt{3}{\sec }^2\theta d\theta$

$L={\int }_0^{\pi /6}\frac{\sqrt{3}{\sec }^2\theta d\theta }{[8{\tan }^2\theta +3{]}^{3/2}}={\int }_0^{\pi /6}\frac{\sqrt{3}{\sec }^2\theta d\theta }{[3{\sec }^2\theta {]}^{3/2}}$

$={\int }_0^{\pi /6}\frac{\sqrt{3}{\sec }^2\theta d\theta }{3\sqrt{3}{\sec }^3\theta }$

$=\frac{1}{3}{\int }_0^{\pi /6}\cos \theta d\theta =\frac{1}{3}\sin \theta {\int }_0^{\pi /6}$

$=\frac{1}{3}[\sin \frac{\pi }{6}-\sin 0]=\frac{1}{3}[\frac{1}{2}-0]=\frac{1}{6}$

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

$\Rightarrow k+5=6k$

$\Rightarrow +5k=+5$

$\Rightarrow k=1$