Answer
$\frac{1}{5}\ln{\frac{7}{2}}$
Work Step by Step
$\frac{1}{x(2x+5)}$ = $\frac{A}{x}+\frac{B}{2x+5}$
$1$ = $A(2x+5)+Bx$
$A$ = $\frac{1}{5}$
$B$ = $-\frac{2}{5}$
$\int{\frac{dx}{x(2x+5)}}$ = $\frac{1}{5}\int{\frac{dx}{x}}$ - $\frac{2}{5}\int{\frac{dx}{2x+5}}$ = $\frac{1}{5}\ln|x|-\frac{1}{5}\ln|2x+5|+C$ = $\frac{1}{5}\ln|\frac{x}{2x+5}|+C$
$\int_1^R{\frac{dx}{x(2x+5)}}$ = $\frac{1}{5}\ln|\frac{x}{2x+5}| |_1^R$ = $\frac{1}{5}\ln|\frac{R}{2R+5}|-\frac{1}{5}\ln{\frac{1}{7}}$
$I$ = $\lim\limits_{R \to {\infty}}$$(\frac{1}{5}\ln|\frac{R}{2R+5}|-\frac{1}{5}\ln{\frac{1}{7}})$ = $\frac{1}{5}\ln{\frac{1}{2}}-\frac{1}{5}\ln{\frac{1}{7}}$ = $\frac{1}{5}\ln{\frac{7}{2}}$
