Answer
$$-\ln |x-2|- \frac{2}{x-2}+\frac{1}{2} \ln \left|x^{2}+4\right|+C$$
Work Step by Step
Given
$$\int \frac{16 d x}{(x-2)^{2}\left(x^{2}+4\right)}$$
Since
\begin{aligned}
\frac{16}{(x-2)^{2}\left(x^{2}+4\right)}&=\frac{A}{(x-2)}+\frac{B}{(x-2)^{2}}+\frac{C x+D}{\left(x^{2}+4\right)}\\
&= \frac{A(x-2)\left(x^{2}+4\right)+B\left(x^{2}+4\right)+(C x+D)(x-2)^{2}}{(x-2)^{2}\left(x^{2}+4\right)}\\
16&=A(x-2)\left(x^{2}+4\right)+B\left(x^{2}+4\right)+(C x+D)(x-2)^{2}
\end{aligned}
Then
\begin{align*}
\text{at } x&=2 \ \ \ \ \to B=2 \\
\text{ coefficient of } x^3&\to\ \ \ \ \to A+C=0 \\
\text{ coefficient of } x^2&\to\ \ \ \ \to 2 A+4 C-D=2 \\
\text{ coefficient of } x &\to\ \ \ \ \to D=0
\end{align*}
Hence $ A=-1,\ \ C=1$ and
\begin{aligned}
\int \frac{16}{(x-2)^{2}\left(x^{2}+4\right)} d x &=-\int \frac{1}{(x-2)} d x+2 \int(x+2)^{-2} d x+\int \frac{ x}{\left(x^{2}+4\right)} d x \\
&=-\int \frac{1}{(x-2)} d x+2 \int(x+2)^{-2} d x+\frac{1}{2} \int \frac{2 x}{\left(x^{2}+4\right)} d x \\
&=-\ln |x-2|+2 \cdot \frac{(x-2)^{-1}}{(-1)}+\frac{1}{2} \ln \left|x^{2}+4\right|+C\\
&=-\ln |x-2|- \frac{2}{x-2}+\frac{1}{2} \ln \left|x^{2}+4\right|+C
\end{aligned}
