1. We need to evaluate

2. First, complete the square in the denominator: $x^2-2x+4=(x-1)^2+3.$ So, $I=\int_1^2 \frac{dx}{\big((x-1)^2+3\big)^{3/2}}.$

3. Substitute $u=x-1 \implies du=dx.$ When $x=1$, $u=0$; when $x=2$, $u=1$. Thus, $I=\int_0^1 \frac{du}{(u^2+3)^{3/2}}.$

4. Use the standard result $\int \frac{du}{(u^2+a^2)^{3/2}}=\frac{u}{a^2\sqrt{u^2+a^2}}+C.$ Here, $a^2=3$. Therefore, $I=\left[\frac{u}{3\sqrt{u^2+3}}\right]_0^1.$

5. Evaluate at the limits: $I=\frac{1}{3\sqrt{4}}-0=\frac{1}{6}.$

6. Given $I=\frac{k}{k+5},$ so $\frac{k}{k+5}=\frac16.$ Cross-multiplying, $6k=k+5$ $5k=5$ $k=1.$

7. Check options:

• A: $1$ ✅
• B: $2$ ❌
• C: $3$ ❌
• D: $4$ ❌

Hence, the correct answer is A.