Half an integral

$\quad f(x) = \dfrac{1}{x^2+4x+4} = \dfrac{1}{(x+2)^2} \\ \quad \displaystyle\int_{1}^{k} f(x) \cdot dx = \dfrac{1}{3}-\dfrac{1}{k+2} \\ \quad \dfrac{1}{2}\displaystyle\int_{1}^{\infty} f(x) \cdot dx = \dfrac{1}{2}\left( \dfrac{1}{3}-\dfrac{1}{\infty+2} \right) = \dfrac{1}{6} \\ \quad \dfrac{1}{6} = \dfrac{1}{3}-\dfrac{1}{k+2} \\ \quad \boxed{k=4}$

$\int_{1}^{\infty}(f(x)dx)=\lim_{c \to \infty} \int_{1}^{c}(\frac{1}{x^2+4x+4}dx)=\lim_{c \to \infty} \int_{1}^{c}(\frac{1}{(x+2)^2}dx) =\lim_{c \to \infty}(-\frac{1}{c+2}+\frac{1}{1+2})=\frac{1}{3}$

We get

$\int_{1}^k(f(x)dx)=\frac{1}{2} \times \frac{1}{3}$

$\int_{1}^k(\frac{1}{(x+2)^2}dx)=\frac{1}{6}$

$-\frac{1}{k+2}+\frac{1}{3}=\frac{1}{6}$

$\frac{1}{k+2}=\frac{1}{6}$

$k+2=6$

$k=\boxed{4}$