Answer
$\frac{1}{6}-\frac{1}{4}\ln{\frac{5}{3}}$
Work Step by Step
$\frac{1}{(x+3)(x+1)^{2}}$ = $\frac{A}{x+3}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$
$1$ = $A(x+1)^{2}+B(x+1)(x+3)+C(x+3)$
$A$ = $\frac{1}{4}$
$B$ = $-\frac{1}{4}$
$C$ = $\frac{1}{2}$
$\int{\frac{dx}{(x+3)(x+1)^{2}}}$ = $\frac{1}{4}\int{\frac{dx}{x+3}}-\frac{1}{4}\int{\frac{dx}{x+1}}+\frac{1}{2}\int{\frac{dx}{(x+1)^{2}}}$ = $\frac{1}{4}\ln|x+3|-\frac{1}{4}\ln|x+1|-\frac{1}{2(x+1)}+C$ = $\frac{1}{4}\ln|\frac{x+3}{x+1}|-\frac{1}{2(x+1)}+C$
$\int_2^R{\frac{dx}{(x+3)(x+1)^{2}}}$ = $\frac{1}{4}\ln|\frac{x+3}{x+1}|-\frac{1}{2(x+1)}|_2^R$ = $\frac{1}{4}\ln|\frac{R+3}{R+1}|-\frac{1}{2(R+1)}-\frac{1}{4}\ln{\frac{5}{3}}+\frac{1}{6}$
$I$ = $\lim\limits_{R \to {\infty}}$$(\frac{1}{4}\ln|\frac{R+3}{R+1}|-\frac{1}{2(R+1)}-\frac{1}{4}\ln{\frac{5}{3}}+\frac{1}{6})$ = $\frac{1}{6}-\frac{1}{4}\ln{\frac{5}{3}}$
