Question

If the integral \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}\], then find the value of \[k\].
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

Hint: In this question, in order to the value of \[k\] given that integral \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}\], we have to first simplify the integrand by substituting \[x-1=\sqrt{3}\tan y\], then in the give integral the lower limit and upper limit of the variable \[x\] should be changed \[y\] by putting the value \[x=1\] and \[x=2\] in \[x-1=\sqrt{3}\tan y\] to find the respective lower limit and upper limit of the variable when we are changing the variable from \[x\] to \[y\]. Also we have to determine the value of \[dx\] in terms of the variable \[y\] and \[dy\]. We will then evaluate the simplified integral in terms of variable \[y\].

Complete step by step answer:
Let \[I\] denote the integral \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}\].
That is, let \[I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}...........(1)\].
Now first we will factorise the expression \[{{x}^{2}}-2x+4\] by splitting the terms.
Then we have
\[\begin{align}
  & {{x}^{2}}-2x+4={{x}^{2}}-2x+1+3 \\
 & ={{\left( x-1 \right)}^{2}}+3..........(2)
\end{align}\]
Now on substituting the value of equation (2) in equation (1), we get
\[I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+3 \right)}^{\dfrac{3}{2}}}}}...........(3)\]
Now let us suppose that \[x-1=\sqrt{3}\tan y\].
Now on differentiate \[x-1=\sqrt{3}\tan y\] where differentiation of \[\tan y\] is equals to \[{{\sec }^{2}}y\] in order to determine the value of \[dx\] in terms of the variable \[y\] and \[dy\], we will get
\[dx=\sqrt{3}{{\sec }^{2}}ydy......(4)\].
Now we will evaluate the lower limit of the integral \[I\] by the value \[x=1\] in \[x-1=\sqrt{3}\tan y\].
Putting the value the value \[x=1\] in \[x-1=\sqrt{3}\tan y\] , we get
\[\begin{align}
  & 1-1=\sqrt{3}\tan y \\
 & \Rightarrow 0=\sqrt{3}\tan y \\
 & \Rightarrow 0=\tan y \\
\end{align}\]
Since we have \[\tan y=0\] when \[y=0\], thus the lower limit of the variable \[y\] is \[0\].
We will now calculate the upper limit of the integral \[I\] by the value \[x=2\] in \[x-1=\sqrt{3}\tan y\].
Putting the value the value \[x=2\] in \[x-1=\sqrt{3}\tan y\] , we get
\[\begin{align}
  & 2-1=\sqrt{3}\tan y \\
 & \Rightarrow 1=\sqrt{3}\tan y \\
 & \Rightarrow \tan y=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Since we have \[\tan y=\dfrac{1}{\sqrt{3}}\] when \[y=\dfrac{\pi }{6}\], thus the lower limit of the variable \[y\] is \[\dfrac{\pi }{6}\].
Now on equation (4) in equation (3) and by changing the lower limit and upper limit of variable \[y\] in equation (3), we will get
\[\begin{align}
  & I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+3 \right)}^{\dfrac{3}{2}}}}} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left( 3{{\tan }^{2}}y+3 \right)}^{\dfrac{3}{2}}}}dy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3\left( {{\tan }^{2}}y+1 \right) \right]}^{\dfrac{3}{2}}}}dy}
\end{align}\]
Using \[1+{{\tan }^{2}}y={{\sec }^{2}}y\] in the above integral, we get
\[\begin{align}
  & I==\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3\left( {{\tan }^{2}}y+1 \right) \right]}^{\dfrac{3}{2}}}}dy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3{{\sec }^{2}}y \right]}^{\dfrac{3}{2}}}}dy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{3\sqrt{3}{{\sec }^{3}}y}dy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3\sec y}dy}
\end{align}\]
 We will now use the identity \[\cos y=\dfrac{1}{\sec y}\] in the above integral.

That is on substituting \[\cos y=\dfrac{1}{\sec y}\] in the above integral, we will have
\[\begin{align}
  & I=\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3\sec y}dy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3}\cos ydy} \\
 & =\dfrac{1}{3}\int\limits_{0}^{\dfrac{\pi }{6}}{\cos ydy} \\
 & =\dfrac{1}{3}\left[ \sin y \right]_{0}^{\dfrac{\pi }{6}} \\
 & =\dfrac{1}{3}\left[ \sin \dfrac{\pi }{6}-\sin 0 \right] \\
 & =\dfrac{1}{3}\left[ \dfrac{1}{2}-0 \right] \\
 & =\dfrac{1}{6}
\end{align}\]
Therefore we have \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{1}{6}\].
But since it is given that \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}\], therefore we have
\[\dfrac{k}{k+5}=\dfrac{1}{6}\]
By cross multiplication of the above equation, we get
\[\begin{align}
  & 6k=k+5 \\
 & \Rightarrow 6k-k=5 \\
 & \Rightarrow 5k=5 \\
 & \Rightarrow k=1
\end{align}\]
Therefore we have that the value of \[k\] is equal to 1.

So, the correct answer is “Option A”.

Note: In this problem, while evaluating the integrals \[\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}\] we are changing the integral in terms of variable \[y\] using \[x-1=\sqrt{3}\tan y\]. Please take care of the fact that while evaluating the integral in terms of variable \[y\] we have to change everything from variable \[x\] to \[y\] including the lower limit and the upper limit of the integral.