integration - thoda hatke

this sum can be done by using the properties of integration
let i be the given integration
the value of I does'nt change when x is replaced by( 0+1-x)
0∫1dx/[(5+2x-2x²)(1+e^(2-4x))]
so this is equal to
_0∫1 dx/[(5+2x-2x²)(1+e^-(2-4x))]
so
I=0∫1e^(2-4x)dx/[(5+2x-2x²)(1+e^(2-4x))]
by adding both
2I=I1
so the answer is none

REplace x by 1-x in I......
Then add original I and this I..........
This will eliminate e^nonsense from the denominator.....
I think so [12]
Didn't do it as yet......... or in the words of our Minister.......Tukka lagaya [3]

but how??
try the method I have posted

chazi u did a chhhhhhhhota sa calculation mistake ...

or else wonderful..

but u cud have got a negative marking..

its [a].

bull shit!!!!
I under stood!!!
damn !!!!!!!!!!!!!!!!!!!!

did this ... straight-cut method...

jus gave ..

thanks