Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

${\int }_1^2\frac{dx}{(x^2-2x+4{)}^{\frac{3}{2}}}={\int }_1^2\frac{dx}{\left[\right(x-1{)}^2+3{]}^{\frac{3}{2}}}$

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

when $x=1\Rightarrow \theta =0,$

when $x=2$, $\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$

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

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

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

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

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

$=\frac{1}{3}\sin \theta {|}_0^{\frac{\pi }{6}}$

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

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

$k+5=6k$

$5=5k$

$k=1$.