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dx-5e-2x-4e-x-1-




Question Number 102461 by bemath last updated on 09/Jul/20
∫ (dx/( (√(5e^(2x) +4e^x +1)))) =?
dx5e2x+4ex+1=?
Answered by PRITHWISH SEN 2 last updated on 09/Jul/20
∫(dx/(e^x (√((1/e^(2x) )+(4/e^x )+5))))   put (1/e^x ) = t⇒(dx/e^x ) = −dt  = −∫(dt/( (√(t^2 +4t+5)) )) = −∫(dt/( (√((t+2)^2 +1))))  =−ln∣(t+2)+(√(t^2 +4t+5))∣ +C  =ln∣(e^x /(2e^x +1+(√(5e^(2x) +4e^x +1))))∣+C  please check.
dxex1e2x+4ex+5put1ex=tdxex=dt=dtt2+4t+5=dt(t+2)2+1=ln(t+2)+t2+4t+5+C=lnex2ex+1+5e2x+4ex+1+Cpleasecheck.

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