Integral from 0 to ln 3 of #e^(3x)/(e^(6x)+5)# . Help please?

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Answer 1 · 2018-05-02T08:58:36

See process below

#I=inte^(3x)/(e^(6x)+5)dx#

Change of variable #t=e^(3x)#, then #dt=3e^(3x)dx=3tdx#

With this change we have

#I=intcancelt/(t^2+5)·1/(3cancelt)dt=1/3int1/(t^2+5)#

Other change of variable #t=sqrt5tantheta#. Then #dt=sqrt5sec^2theta# and integral

#I=1/3int1/(5tan^2theta+5)·sqrt5sec^2thetad theta=#

#=1/3·sqrt5/5int(cancelsec^2thetad theta)/(cancelsec^2theta)=#

#=sqrt5/15intd theta=sqrt5/15theta=sqrt5/15arctan(t/sqrt5)=#

#=sqrt5/15arctan(e^(3x)/sqrt5)=F(x)#

#F(ln3)-F(0)=sqrt5/15arctan(27/sqrt5)-sqrt5/15arctan(1/sqrt5)#