Question Number 201925 by Calculusboy last updated on 15/Dec/23
Commented by Calculusboy last updated on 17/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by Frix last updated on 16/Dec/23
$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{tan}\:{x}\:+\mathrm{tan}^{\mathrm{2}} \:{x}\:−\mathrm{tan}^{\mathrm{3}} \:{x}\:+\mathrm{tan}^{\mathrm{4}} \:{x}}{dx} \\ $$$$\mathrm{Use}\:{t}=\mathrm{tan}\:{x}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{2}\sqrt{\mathrm{5}\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{25}}\pi \\ $$