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2 solutions
u have to just find the indefinite integral which are just a few addition's and mul. consider
2x(1-x)
= -0.25 ( -8x(1-x) )
= -0.25 ( -2+2-8x(1-x) )
= -0.25 ( -2+2(1-4x(1-x) ) )
Thus
2x(1-x) e^2x(1-x)
=e^2x(1-x) [-0.25 ( -2+2(1-4x(1-x) ) )]
=-0.25 e^2x(1-x) [ -2+2(1-4x(1-x) ) ]
=using product rule
=-0.25 d/dx { (1-2x) e^2x(1-x) }
just integrate and get
I = -0.25 * (1-2x)e^2x(1-x)
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u mean this
2 x ( 1 − x )
= − 0 . 2 5 ( − 8 x ( 1 − x ) )
= − 0 . 2 5 ( − 2 + 2 − 8 x ( 1 − x ) )
= − 0 . 2 5 ( − 2 + 2 ( 1 − 4 x ( 1 − x ) ) ) Thus
2 x ( 1 − x ) e 2 x ( 1 − x )
= e 2 x ( 1 − x ) [ − 0 . 2 5 ( − 2 + 2 ( 1 − 4 x ( 1 − x ) ) ) ]
= − 0 . 2 5 e 2 x ( 1 − x ) [ − 2 + 2 ( 1 − 4 x ( 1 − x ) ) ] =using product rule
= − 0 . 2 5 d x d ( 1 − 2 x ) e 2 x ( 1 − x )
just integrate and get
I = − 0 . 2 5 ∗ ( 1 − 2 x ) e 2 x ( 1 − x )
I used a series to approximate the answer as 3 1 + 3 ⋅ 5 1 + 3 ⋅ 5 ⋅ 7 1 + . . . , which I'm not even sure is correct; I could've made a mistake. But, obviously, your way is better. I would have written it like this, however: 2 x ( 1 − x ) e 2 x ( 1 − x ) = − 0 . 2 5 ( 2 ( 4 x ( x − 1 ) e 2 x ( 1 − x ) ) ) = − 0 . 2 5 ( 2 ( 4 x 2 − 4 x ) e 2 x ( 1 − x ) ) = − 0 . 2 5 ( 2 ( ( 2 x − 1 ) 2 − 1 ) e 2 x ( 1 − x ) ) = − 0 . 2 5 ( 2 ( 1 − 2 x ) 2 e 2 x ( 1 − x ) − 2 e 2 x ( 1 − x ) ) = − 0 . 2 5 ( ( 1 − 2 x ) d ( 2 x ( 1 − x ) ) d [ e 2 x ( 1 − x ) ] ( 2 − 4 x ) − 2 e 2 x ( 1 − x ) ) = − 0 . 2 5 ( ( 1 − 2 x ) d ( 2 x ( 1 − x ) ) d [ e 2 x ( 1 − x ) ] d x d ( 2 x ( 1 − x ) ) + d x d [ 1 − 2 x ] e 2 x ( 1 − x ) ) = − 0 . 2 5 ( ( 1 − 2 x ) d x d [ e 2 x ( 1 − x ) ] + d x d [ 1 − 2 x ] e 2 x ( 1 − x ) ) = − 0 . 2 5 d x d [ ( 1 − 2 x ) e 2 x ( 1 − x ) ]
Set 2 x ( 1 − x ) e 2 x ( 1 − x ) = d x d [ ( a + b x ) e 2 x ( 1 − x ) ]
= b e 2 x ( 1 − x ) + ( a + b x ) e 2 x ( 1 − x ) ( 2 − 4 x )
= ( 2 a + b + ( − 4 a + 2 b ) x − 4 b x 2 ) e 2 x ( 1 − x ) .
Equating coefficients results in
2 a + b = 0
− 4 a + 2 b = 2
− 4 b = − 2
There is a unique solution to these equations, namely, ( a , b ) = ( − 4 1 , 2 1 ) .
Thus, ∫ 2 x ( 1 − x ) e 2 x ( 1 − x ) d x = ( − 4 1 + 2 1 x ) e 2 x ( 1 − x ) + C .
( − 4 1 + 2 1 x ) e 2 x ( 1 − x ) ∣ ∣ ∣ 0 1 = ( − 4 1 + 2 1 ) e 0 − ( − 4 1 ) e 0 = 2 1 .
incredible mind Do post a solution for this, frankly typing XD i used wolfram alpha.