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# ∫dx/[1+3ex+2e2x] ∫[ex(1-x2)/(1+x2)]dx 7 years ago

Dearc Student ,

Integral (dx/(1+3e^x+2e^(2x)))

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

7 years ago

Dear Student ,

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