Rinkoo Gupta
Last Activity: 11 Years ago
Dear Student ,
Pls type one question in one linkso that we can help you
better.
Integral (dx/(1+3e^x+2e^(2x)))
Let e^x=t so that e^x dx=dt=>dx=dt/e^x=>dx=dt/t,then we get
Integral(dt/t(1+3t+2t^2)
Now Using Partial Fraction ,we get
Integral [1/t – (2t+3)/(1+3t+2t^2)]dt
=integral dt/t- integral (2t+3)/(1+3t+2t^2) dt
=logt- integral[{(4t+3)/2
+3/2}/(2t^2+3t+1)]dt
=logt-1/2integral{(4t+3)/(2t^2+3t+1)}dt -3/2integral
dt/(2t^2+3t+1)
=logt -1/2log(2t^2+3t+1) -3/4 integral(dt/(t^2+3t/2+1/2+9/16
-9/16))
=logt -1/2log(2t^2+3t+1)-3/4integral dt/((t+3/4)^2-(1/4)^2)
=logt-1/2log(2t^2+3t+1)-(3/4).(1/2(1/4))log mod((t+3/4-1/4)/(t+3/4+1/4))+C
=logt-(1/2)log(2t^2+3t+1)-3/2logmod(t+1/2)/(t+1))+C
=log(e^x)-(1/2)log(2t^2+3t+1) -3/2logmod((t+1/2)/(t+1))+c
=x-(1/2)log(2e^(2x)+3e^x+1))-3/2logmod((e^x+1/2)/(e^x+1)) +C Ans
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty