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Question Number 142365 by mathmax by abdo last updated on 30/May/21
calculate∫1+ex+e2xdx
Answered by MJS_new last updated on 05/Jun/21
∫e2x+ex+1dx=[t=ex+12→dx=dtex]=∫4t2+32t−1dt=[u=2t+4t2+33→dt=4t2+32udu]=334∫(u2+1)2u2(u−3)(3u+3)du==∫(34−34u2+12u+1u−3−33u+3)du==3u4+34u+lnu2+ln(u−3)−ln(3u+3)==3(u2+1)4u+12lnu(u−3)2(3u+3)2=...=e2x+ex+1−x+12ln(2e3x−3e2x−8+2(e2x−2ex+4)e2x+ex+1)+C
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