Evaluate the integral of primitive $\frac{1}{x(x+2)}$ I am doing this integral by example:
$$\int_1^\infty\frac{1}{x(x+2)}\ dx.$$
The example in the book, starts with $\dfrac{1}{x(x+2)} \leq \dfrac{1}{x^2}.$
Why is this important? Why does the example in the book start like that? 
Thanks.

If $\dfrac{1}{x(x+2)}\leq\dfrac{1}{x^2}$, then 
$$\int^\infty_1\dfrac{1}{x(x+2)}dx\leq\int^\infty_1\dfrac{1}{x^2}dx=\left.-\dfrac{1}{x}\right|^\infty_1=1$$
i.e. $\displaystyle\int^\infty_1\dfrac{1}{x(x+2)}dx\leq1$. This shows convergence of the integral.
Now, split $\dfrac{1}{x(x+2)}$ as a partial fraction:
$$\dfrac{1}{x(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+2}=\dfrac{(A+B)x+2A}{x(x+2)}\\
\implies (A+B)x+2A=1.$$
Solve for $A$ and $B$, then substitute them into $\dfrac{1}{x(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+2}$, and integrate it term by term.
