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## $\mathrm{Let}I=\int \frac{1}{\left(\mathrm{sin}x-2\mathrm{cos}x\right)\left(2\mathrm{sin}x+\mathrm{cos}x\right)}dx\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}⇒I=\int \frac{{\mathrm{sec}}^{2}x}{\left(\frac{\mathrm{sin}x-2\mathrm{cos}x}{\mathrm{cos}x}\right)×\left(\frac{2\mathrm{sin}x+\mathrm{cos}x}{\mathrm{cos}x}\right)}dx\phantom{\rule{0ex}{0ex}}=\int \frac{{\mathrm{sec}}^{2}x}{\left(\mathrm{tan}x-2\right)\left(2\mathrm{tan}x+1\right)}dx\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{tan}x=t\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sec}}^{2}xdx=dt\phantom{\rule{0ex}{0ex}}\therefore I=\int \frac{dt}{\left(t-2\right)\left(2t+1\right)}\phantom{\rule{0ex}{0ex}}=\int \frac{dt}{2{t}^{2}+t-4t-2}\phantom{\rule{0ex}{0ex}}=\int \frac{dt}{2{t}^{2}-3t-2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \frac{dt}{{t}^{2}-\frac{3}{2}t-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \frac{dt}{{t}^{2}-\frac{3}{2}t+{\left(\frac{3}{4}\right)}^{2}-{\left(\frac{3}{4}\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \frac{dt}{{\left(t-\frac{3}{4}\right)}^{2}-\frac{9}{16}-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \frac{dt}{{\left(t-\frac{3}{4}\right)}^{2}-{\left(\frac{5}{4}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{1}{2×\frac{5}{4}}\mathrm{log}\left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\mathrm{ln}\left|\frac{t-2}{t+\frac{1}{2}}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\mathrm{ln}\left|\frac{{\left(t-2\right)}^{2}}{2t+1}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\mathrm{ln}\left|\frac{t-2}{2t+1}\right|+\frac{1}{5}\mathrm{ln}\left(2\right)+C\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\mathrm{ln}\left|\frac{t-2}{2t+1}\right|+C\text{'}\mathrm{where}C\text{'}=C+\frac{1}{5}\mathrm{ln}\left(2\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\mathrm{ln}\left|\frac{\mathrm{tan}x-2}{2\mathrm{tan}x+1}\right|+C$

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