`inte^(2x)((sin4x-2)/(1-cos4x))dx`

Explanation

Putting 2x=t anddx `=(1)/(2)dt`, we get
`I=(1)/(2)inte^(t)((sin2t-2)/(1-cos2t))dt=(1)/(2)inte^(t)((2sintcos-2)/(2sin^(2)))dt`
`=(1)/(2)inte^(t){(sintcost-1)/(sin^(2)t)}dt=(1)/(2)inte^(t)(cot t-"cosec"^(2)t)dt`
`=(1)/(2)inte^(t){f(t)+f'(t)}dt,"where"f(t)=cott`
`=(1)/(2)e^(t)*f(t)+C=(1)/(2)e^(t)cott+C=(1)/(2)e^(2x)cot2x+C`.