Question

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

Answer 1

`(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`