Question

If $\underset{1}{\overset{2}{\int }}\frac{dx}{{\left(x^2-2x+4\right)}^{\frac{3}{2}}}=\frac{k}{k+5^{\prime }}$ , then $k$ is equal to :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

$\underset{1}{\overset{2}{\int }}\frac{dx}{(x-1)+3{)}^{3/2}}$
$x-1=\sqrt{3}\tan Q$
$=\sqrt{3}{\sec }^{2}Q$
$\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{\sqrt{3}\sec dQ}{3\sqrt{3}\sec .3Q}$
$=\frac{1}{3}\underset{0}{\overset{\frac{\pi }{6}}{\int }}\cos esQ=\frac{1}{3}(\tan Q{)}_0^{\pi /6}$
$=\frac{1}{6}=\frac{k}{k+5}=k+5=k$
$=k=1$