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1.44 Integration using partial fractions

To show that

\int_{2}^{\infty}{{1}\over{x^{2}(2x+1)}}\,dx=2\log{{4}\over{5}}+{{1}\over{2}}.

Solution. Here the partial fractions decomposition is

{{1}\over{x^{2}(2x+1)}}={{Ax+B}\over{x^{2}}}+{{C}\over{2x+1}}

where $(Ax+B)(2x+1)+Cx^{2}=1$ , so $(2A+C)x^{2}+(A+2B)x+B=1$ . Thus $C=-2A$ , $A=-2B$ and $B=1$ , so we obtain:

{{{1}\over{x^{2}(2x+1)}}=\frac{1}{x^{2}}-\frac{2}{x}+\frac{4}{2x+1}}{=\frac{1}% {x^{2}}-\frac{2}{x}+\frac{2}{x+\frac{1}{2}}.}