$int\frac{1}{(x - 2)(x^{2}+4)}dx$

Explanation:

Step1: Use partial - fraction decomposition

Let $\frac{1}{(x - 2)(x^{2}+4)}=\frac{A}{x - 2}+\frac{Bx + C}{x^{2}+4}$. Then $1=A(x^{2}+4)+(Bx + C)(x - 2)$. Set $x = 2$, we get $1=A(4 + 4)$, so $A=\frac{1}{8}$. Expand the right - hand side: $1=Ax^{2}+4A + Bx^{2}-2Bx + Cx-2C=(A + B)x^{2}+(-2B + C)x+(4A-2C)$. Comparing the coefficients of $x^{2}$: $0=A + B$, since $A=\frac{1}{8}$, then $B=-\frac{1}{8}$. Comparing the coefficients of $x$: $0=-2B + C$, substituting $B = -\frac{1}{8}$, we get $C=-\frac{1}{4}$. So $\frac{1}{(x - 2)(x^{2}+4)}=\frac{1}{8(x - 2)}-\frac{x + 2}{8(x^{2}+4)}$.

Step2: Integrate term - by - term

$\int\frac{1}{(x - 2)(x^{2}+4)}dx=\frac{1}{8}\int\frac{1}{x - 2}dx-\frac{1}{8}\int\frac{x}{x^{2}+4}dx-\frac{1}{4}\int\frac{1}{x^{2}+4}dx$. For $\int\frac{1}{x - 2}dx=\ln|x - 2|+C_1$. For $\int\frac{x}{x^{2}+4}dx$, let $u=x^{2}+4$, $du = 2xdx$, then $\int\frac{x}{x^{2}+4}dx=\frac{1}{2}\ln|x^{2}+4|+C_2$. For $\int\frac{1}{x^{2}+4}dx=\int\frac{1}{4(\frac{x^{2}}{4}+1)}dx=\frac{1}{2}\arctan(\frac{x}{2})+C_3$. So $\int\frac{1}{(x - 2)(x^{2}+4)}dx=\frac{1}{8}\ln|x - 2|-\frac{1}{16}\ln(x^{2}+4)-\frac{1}{8}\arctan(\frac{x}{2})+C$.

Answer:

$\frac{1}{8}\ln|x - 2|-\frac{1}{16}\ln(x^{2}+4)-\frac{1}{8}\arctan(\frac{x}{2})+C$