Mathematics

# $\displaystyle{\int}_{1}^{2}\dfrac{dx}{\left(x^{2}-2x+4\right)^{\tfrac{3}{2}}}=\dfrac{k}{k+5}$, then $k$ is equal to

$1$

##### SOLUTION
$\displaystyle \int_1^2 \dfrac{dx}{(x^2 - 2x + 1)^{2/3}} = \int_1^2 \dfrac{dx}{[(x - 1)^2 + 3]^{3/2}}$
Substitute $x - 1 = \sqrt{3} \tan \theta$          when $x = 2$,
when $x = 1, \theta = 0$                          $\theta = \tan^{-1} \left(\dfrac{1}{\sqrt{3}} \right) = \dfrac{\pi}{6}$
$dx = \sqrt{3} \, \sec^2 \theta \, d \theta$
$L = \displaystyle \int_0^{\pi/6} \dfrac{\sqrt{3} \sec^2 \theta d \theta}{[8 \tan^2 \theta + 3]^{3/2}} = \int_0^{\pi/6} \dfrac{\sqrt{3} \sec^2 \theta \, d \theta}{[3 \sec^2 \theta]^{3/2}}$
$= \displaystyle \int_0^{\pi/6} \dfrac{\sqrt{3} \sec^2 \theta \, d \theta}{3 \sqrt{3} \sec^3 \theta}$
$= \dfrac{1}{3} \displaystyle \int_0^{\pi/6} \cos \theta d \theta = \dfrac{1}{3} \sin \theta \int_0^{\pi/6}$
$= \dfrac{1}{3} \left[\sin \dfrac{\pi}{6} - \sin 0\right] = \dfrac{1}{3} \left[\dfrac{1}{2} - 0\right] = \dfrac{1}{6}$
$\dfrac{1}{6} = \dfrac{k}{k + 5}$
$\Rightarrow k + 5 = 6k$
$\Rightarrow +5k = + 5$
$\Rightarrow k = 1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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