Question Number 138403 by tugu last updated on 13/Apr/21
$$\int\frac{{e}^{\mathrm{4}{t}} }{{e}^{\mathrm{2}{t}} +\mathrm{3}{e}^{{t}} +\mathrm{2}}{dt}=? \\ $$
Answered by bemath last updated on 13/Apr/21
$${let}\:{e}^{{t}} \:=\:{u}\:,\:{e}^{\mathrm{2}{t}} +\mathrm{3}{e}^{{t}} +\mathrm{2}={u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{2} \\ $$$$=\left({u}+\mathrm{1}\right)\left({u}+\mathrm{2}\right) \\ $$$${e}^{{t}} \:{dt}\:=\:{du}\: \\ $$$$\Rightarrow{I}=\int\frac{{u}^{\mathrm{3}} }{{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{2}}\:{du} \\ $$