If $$\int_1^2 \frac{dx}{(x^2 - 2x + 4)^{\frac{3}{2}}} = \frac{k}{k+5}$$, then $$k$$ is equal to

The definite integral that has been given is

$$I=\int_{1}^{2}\dfrac{dx}{\left(x^{2}-2x+4\right)^{\dfrac32}}.$$

First the quadratic inside the power is written in its completed-square form:

$$x^{2}-2x+4 \;=\;(x-1)^{2}+3.$$

Put the simple translation

$$t=x-1 \;\Longrightarrow\; dt=dx,\quad x=1\Longrightarrow t=0,\; x=2\Longrightarrow t=1.$$

In terms of the new variable the integral becomes

$$I=\int_{0}^{1}\dfrac{dt}{\left(t^{2}+3\right)^{\dfrac32}}.$$

Next introduce a trigonometric substitution that removes the root in the denominator. Choose

$$t=\sqrt3\,\tan\theta\quad\Longrightarrow\quad dt=\sqrt3\,\sec^{2}\theta\,d\theta.$$

Because $$t$$ runs from $$0$$ to $$1$$, the new variable $$\theta$$ runs from

$$t=0\Longrightarrow\theta=0,\qquad t=1\Longrightarrow 1=\sqrt3\,\tan\theta\Longrightarrow\tan\theta=\dfrac1{\sqrt3}\Longrightarrow\theta=\dfrac{\pi}{6}.$$

Now evaluate the denominator after the substitution:

$$t^{2}+3=\bigl(\sqrt3\,\tan\theta\bigr)^{2}+3=3\tan^{2}\theta+3=3\bigl(1+\tan^{2}\theta\bigr)=3\sec^{2}\theta.$$

Raise this to the three-halves power exactly as it appears in the integrand:

$$\bigl(t^{2}+3\bigr)^{\dfrac32}= \bigl(3\sec^{2}\theta\bigr)^{\dfrac32} = 3^{\dfrac32}\,\bigl(\sec^{2}\theta\bigr)^{\dfrac32} = 3\sqrt3\,\sec^{3}\theta.$$

Substituting everything into the integral, the integrand simplifies step by step:

$$ I =\int_{0}^{\pi/6} \dfrac{\sqrt3\,\sec^{2}\theta\,d\theta} {3\sqrt3\,\sec^{3}\theta} =\int_{0}^{\pi/6}\dfrac{\sqrt3\,\sec^{2}\theta\,d\theta} {3\sqrt3\,\sec^{3}\theta} =\int_{0}^{\pi/6}\dfrac{\sec^{2}\theta}{3\,\sec^{3}\theta}\,d\theta =\int_{0}^{\pi/6}\dfrac{1}{3\,\sec\theta}\,d\theta =\int_{0}^{\pi/6}\dfrac{\cos\theta}{3}\,d\theta. $$

Factor the constant $$\dfrac13$$ out of the integral:

$$I=\dfrac13\int_{0}^{\pi/6}\cos\theta\,d\theta.$$

The antiderivative of $$\cos\theta$$ is $$\sin\theta$$, so

$$I=\dfrac13\Bigl[\sin\theta\Bigr]_{0}^{\pi/6} =\dfrac13\left(\sin\dfrac{\pi}{6}-\sin0\right) =\dfrac13\left(\dfrac12-0\right) =\dfrac13\cdot\dfrac12 =\dfrac16.$$

According to the question this value equals the rational expression $$\dfrac{k}{k+5}$$, so write

$$\dfrac{k}{k+5}=\dfrac16.$$

Cross-multiply every term so that no denominator remains:

$$6k=k+5.$$

Bring all terms involving $$k$$ to one side to isolate the unknown:

$$6k-k=5\quad\Longrightarrow\quad5k=5.$$

Divide by $$5$$ to obtain the numerical value of $$k$$:

$$k=\dfrac{5}{5}=1.$$

From the list of alternatives we see that the number $$1$$ corresponds to Option D.

Hence, the correct answer is Option D.