If $\displaystyle \int_{1}^{2} \frac{dx}{(x^{2} - 2x + 4)^{\tfrac{3}{2}}} = \frac{k}{k+5}$, then $k$ is equal to:

Given:
\(\int_1^2 \frac{dx}{(x^2-2x+4)^{3/2}}=\frac{k}{k+5}\)

Now,
\(x^2-2x+4=(x-1)^2+3\)

Put \(x-1=t\), then \(dx=dt\)

When \(x=1\), \(t=0\)
When \(x=2\), \(t=1\)

So,
\(\int_1^2 \frac{dx}{(x^2-2x+4)^{3/2}}\)
\(=\int_0^1 \frac{dt}{(t^2+3)^{3/2}}\)

Using formula:
\(\int \frac{dt}{(t^2+a^2)^{3/2}}=\frac{t}{a^2\sqrt{t^2+a^2}}\)

Here, \(a^2=3\)

So,
\(\int_0^1 \frac{dt}{(t^2+3)^{3/2}}\)
\(=\left[\frac{t}{3\sqrt{t^2+3}}\right]_0^1\)

\(=\frac{1}{3\sqrt{4}}-0\)

\(=\frac{1}{6}\)

Therefore,
\(\frac{k}{k+5}=\frac{1}{6}\)

\(6k=k+5\)

\(5k=5\)

\(k=1\)

Answer: (a) \(1\)