Evaluate `int_0^1(dx)/((5+2x-2x^2)(1+e^(2-4x))) dx`

Explanation

Let `I=int_(0)^(1)(dx)/((5+2x-2x^(2))(1+e^(2-4x)))`
Also `I=int_(0)^(1)(dx)/([5+2(1-x)-2(1-x)^(2)][1+e^(2-4(1-x))])`
`=int_(0)^(1)(dx)/((5+2x-2x^(2))(1+e^(-2+4x)))`
`=int_(0)^(1)(e^(2-4x)dx)/((5+2x-2x^(2))(e^(2-4x)+1))`
Adding 1 and 2 we get
`2I=int_(0)^(1)((1+e^(2-4x))dx)/((5+2x-2x^(2))(1+e^(2-4x)))`
`int_(0)^(1)(dx)/(5-2(x^(2)-x))=int_(0)^(1)(dx)/(1/2+5-2(x-1/2)^(2))`
`=1/2int_(0)^(1)(dx)/(11/4-(x-1/2)^(2))`
`=1/(4sqrt(11)//2)|log(sqrt(11)//2+x-1/2)/(sqrt(11)//2-(x-1/2))|_(0)^(1)`
`=1/(2sqrt(11))["log" ((sqrt(11))2+1/2)/((sqrt(11))/2-1/2)-"log"((sqrt(11))/2-1/2)/((sqrt(11))/2+1/2)]`
`=1/(2sqrt(11))[2log((sqrt(11)+1)/(sqrt(11)-1))]`
`=1/(sqrt(11))log((sqrt(11)+1)/(sqrt(11)-1))`
`=1/(sqrt(11))"log"(sqrt(11)+1)/(sqrt(11)-1)(sqrt(11)+1)/(sqrt(11)+1)`
`=1/(sqrt(11))"log"((sqrt(11)+1)^(2))/10`