## Calculus (3rd Edition)

$$-\ln |x-2|- \frac{2}{x-2}+\frac{1}{2} \ln \left|x^{2}+4\right|+C$$
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}