U-substitution: integral of e^(24x)/(1+e^(8x))?

1 Answer
ProfLayton
Jan 27, 2018

#(1/16)e^(16x)-(1/8)e^(8x)+(1/8)ln|e^(8x)+1|+C#

Explanation:

Let #u=e^(8x)#
#du=8e^(8x)#

#int(e^(24x))/(1+e^(8x))dx#
#=(1/8)int(e^(16x))/(1+e^(8x))*8e^(8x)dx#
#=(1/8)int(u^2)/(1+u)du#
#=(1/8)int((u^2+u)-(u+1)+1)/(1+u)du#
#=(1/8)intu-1+(1)/(1+u)du#
#=(1/8)((1/2)u^2-u+ln|u+1|)+C#
#=(1/16)e^(16x)-(1/8)e^(8x)+(1/8)ln|e^(8x)+1|+C#

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