Question

If $\int_{1}^{2}{\dfrac{dx}{{{({{x}^{2}}-2x+4)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}$ then k is equal to:
(a) $1$
(b) $2$
(c) $3$
(d) $4$

Answer 1

This is the question related to integration and to solve this question we should know all the formulas related to integration. If there is some linear equation given in the denominator then first that equation should be converted into a perfect square and after that, we can solve our question easily.

Complete step by step answer:
This is the question of integration and we have to use some special properties to solve this question.
In this question, we have to find the value of k and it is given that.
$\int_{1}^{2}{\dfrac{dx}{{{({{x}^{2}}-2x+4)}^{\dfrac{3}{2}}}}=\dfrac{k}{k+5}}$………eq(1)
Let, $I=\int_{1}^{2}{\dfrac{dx}{{{({{x}^{2}}-2x+4)}^{\dfrac{3}{2}}}}}$………eq(2)
To solve this type of question we have to make a perfect square in our denominator. So now we will make the square of ${{x}^{2}}-2x+4$. Which is as follows.
Firstly take the constant digit number on the right-hand side.

& {{x}^{2}}-2x+4=0 \\\ & \Rightarrow {{x}^{2}}-2x=-4 \\\ \end{aligned}$$ Then add some number on the left-hand side of the equation to make it a perfect square and also add that number to the right-hand side to balance the equation. $${{x}^{2}}-2x+1=-4+1$$ $$\begin{aligned} & \Rightarrow {{(x-1)}^{2}}=-3 \\\ & \Rightarrow {{(x-1)}^{2}}+3=0 \\\ \end{aligned}$$ So the equation $${{x}^{2}}-2x+4$$ will be written as $${{(x-1)}^{2}}+3$$. Now on putting this value in eq(2), the following results will be obtained. $$\begin{aligned} & I=\int_{1}^{2}{\dfrac{dx}{{{({{x}^{2}}-2x+4)}^{\dfrac{3}{2}}}}} \\\ & \Rightarrow I=\int_{1}^{2}{\dfrac{dx}{{{({{(x-1)}^{2}}+3)}^{\dfrac{3}{2}}}}} \\\ \end{aligned}$$ ……….eq(3) Now we will put the value of $$x-1$$ as $$\sqrt{3}\tan \theta $$ in the above equation. $$x-1=\sqrt{3}\tan \theta $$…….eq(4) We will differentiate the above equation. $$\begin{aligned} & x-1=\sqrt{3}\tan \theta \\\ & dx=\sqrt{3}{{\sec }^{2}}\theta d\theta \\\ \end{aligned}$$ If the value of $$x-1$$ is changed to $$\sqrt{3}\tan \theta $$ then the limits will also be changed. From eq(4), $$x-1=\sqrt{3}\tan \theta $$ If the value of x will be equal to $$1$$, then in this case $$\begin{aligned} & 1-1=\sqrt{3}\tan \theta \\\ & \Rightarrow 0=\sqrt{3}\tan \theta \\\ & \Rightarrow \tan \theta =0 \\\ \end{aligned}$$ So the value of $$\theta =0$$. If the value of x will be equal to $$2$$, then in this case $$\begin{aligned} & 2-1=\sqrt{3}\tan \theta \\\ & \Rightarrow 1=\sqrt{3}\tan \theta \\\ & \Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}} \\\ & \Rightarrow \tan \theta =\tan \dfrac{\pi }{6} \\\ \end{aligned}$$ So the value of $$\theta =\dfrac{\pi }{6}$$. On putting the values and limits in eq(1), the following results will be obtained. $$I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}\theta d\theta }{{{({{(\sqrt{3}\tan \theta )}^{2}}+3)}^{\dfrac{3}{2}}}}}$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}\theta d\theta }{{{(3{{\tan }^{2}}\theta +3)}^{\dfrac{3}{2}}}}}$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}\theta d\theta }{{{(3(1+{{\tan }^{2}}\theta ))}^{\dfrac{3}{2}}}}}$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}\theta d\theta }{{{(3{{\sec }^{2}}\theta )}^{\dfrac{3}{2}}}}}$$ $$\begin{aligned} & \Rightarrow I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}\theta d\theta }{3\sqrt{3}{{\sec }^{3}}\theta }} \\\ & \Rightarrow I=\int_{0}^{\dfrac{\pi }{6}}{\dfrac{d\theta }{3\sec \theta }} \\\ \end{aligned}$$ $$\Rightarrow I=\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{6}}{\cos \theta }d\theta $$ $$\Rightarrow I=\dfrac{1}{3}\left[ \sin \theta \right]_{0}^{\dfrac{\pi }{6}}$$ $$\Rightarrow I=\dfrac{1}{3}\left[ \sin \dfrac{\pi }{6}-\sin 0 \right]$$……..eq(5) We know that the value of $$\sin \dfrac{\pi }{6}=\dfrac{1}{2}$$ $$\sin 0{}^\circ =0$$ On putting these values in eq(5), we get the following results. $$\begin{aligned} & I=\dfrac{1}{3}\left[ \dfrac{1}{2} \right] \\\ & \Rightarrow I=\dfrac{1}{6} \\\ \end{aligned}$$ On putting values in eq(1), we get the following results. $$\dfrac{1}{6}=\dfrac{k}{k+5}$$ $$\begin{aligned} & k+5=6k \\\ & \Rightarrow 5k=5 \\\ & \Rightarrow k=1 \\\ \end{aligned}$$ So the value of k will be $$1$$. **So, the correct answer is “Option a”.** **Note:** The derivative of a constant function always gives the result zero. There are two kinds of integral. The first one is a definite integral and the second one is an indefinite integral. In definite integral, limit will be given to us but in indefinite integral, no limits will be given to us. The reverse function of integral is known as the derivative.