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1.45 Integration using partial fractions continued

We can now calculate the integral:

\int_{2}^{R}{{1}\over{x^{2}(2x+1)}}\,dx=\int_{2}^{R}\left(\frac{1}{x^{2}}-% \frac{2}{x}+\frac{2}{x+\frac{1}{2}}\right)\,dx

={\left[2\log(x+\frac{1}{2})-2\log x-\frac{1}{x}\right]_{2}^{R}}\,{=\left[2% \log\left(\frac{x+\frac{1}{2}}{x}\right)-\frac{1}{x}\right]_{2}^{R}}

{=2\log\left(\frac{R+\frac{1}{2}}{R}\right)-\frac{1}{R}-2\log\frac{5}{4}+\frac% {1}{2}.}

Now ${\frac{R+\frac{1}{2}}{R}=1+\frac{1}{2R}\rightarrow 1}$ as $R\rightarrow\infty$ , and $\frac{1}{R}\rightarrow 0$ as $R\rightarrow\infty$ , so the integral converges to $2\log\frac{4}{5}+\frac{1}{2}$ .