STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

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

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

$\int_{0}^{\pi / 6} \frac{\sqrt{3} \sec ^{2} \theta d \theta}{\left(\sqrt{3} \tan \theta^{2}+(\sqrt{3})^{2}\right)^{3 / 2}}$

$=\frac{1}{3} \int_{0}^{\pi / 6} \frac{\sec ^{2} \theta}{\sec ^{3} \theta} d \theta$

$=\frac{1}{3} \int_{0}^{\pi / 6} \cos \theta d \theta$

$=\frac{1}{3}[\sin \theta]^{6}$

$=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}$

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

$ = \boxed{k = 1}$