\(\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5}\), then \(\mathrm{k}\) has the value

\(\text { Let } \begin{aligned}
\mathrm{I} & =\int_1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}} \\
& =\int_1^2 \frac{\mathrm{d} x}{\left[(x-1)^2+3\right]^{\frac{3}{2}}}
\end{aligned}\)
Put \(x-1=\sqrt{3} \tan \theta\)
\(\mathrm{d} x=\sqrt{3} \sec ^2 \theta \mathrm{d} \theta\)
When \(x=1, \theta=0\)
When \(x=2, \theta=\frac{\pi}{6}\)
\(\begin{aligned} & \therefore \quad I=\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 \theta}{\left[3 \tan ^2 \theta+3\right]^{\frac{3}{2}}} d \theta \\ & =\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 \theta}{\left[3\left(1+\tan ^2 \theta\right)\right]^{\frac{3}{2}}} \\ & =\int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec ^2 \theta}{3 \cdot \sqrt{3}\left(\sec ^2 \theta\right)^{\frac{3}{2}}} \\ & =\int_0^{\frac{\pi}{6}} \frac{1}{3} \cdot \frac{\sec ^2 \theta}{\sec ^3 \theta} \\ & =\frac{1}{3} \int_{\infty}^{\frac{\pi}{6}} \cos \theta \\ & =\frac{1}{3}[\sin \theta]_0^{\frac{\pi}{6}} \\ & I=\frac{1}{3}\left[\sin \frac{\pi}{6}-\sin 0\right] \\ & I=\frac{1}{6} \\ & \therefore \quad \frac{k}{k+5}=\frac{1}{6} \\ & 6 k=k+5 \\ & \therefore \quad \mathrm{k}=1 \\ & \end{aligned}\)