Let I = ∫ ex(

sin4x−4

1−cos4x

) dx
= ∫ ex(

2sin2xcos2x−4

2sin2(2x)

) dx [Using, sin2x 2sinx . cosx and 2 sin2 x = 1 - cos(2x)
= ∫ ex(

2(sin(2x)cos(2x)−4

2sin2x

) dx = ∫ ex(

sin(2x)cos(2x)

sin22x

−

2

sin22x

) dx
= ∫ ex 9cot (2x) - 2 cosec2 2x) dx
Now, let f(x) = cot (2x) then f’(x) = -2 cosec2 2x
I = ∫ ex (f (x) + f' (x)) dx
So, I = ex f(x) + C = ex cot 2x + C , where C is a constant
Therefore, ∫ex(

sin4x−4

1−cos4x

) dx = ex cot (2x) + C
OR
∫

1−x2

x(1−2x)

dx
Here

1−x2

x(1−2x)

is an improper rational fraction
Reducing it to proper rational fraction gives

1−x2

x(1−2x)

=

1

2

+

1

2

(

2−x

x(1−2x)

... (i)
Now, let

2−x

x(1−2x)

=

A

x

+

B

(1−2x)

⇒

2−x

x(1−2x)

=

A(1−2x)+Bx

x(1−2x)

⇒ 2 - x = A - x (2A - B)
Equating the coefficients we get ,A = 2 and B = 3
So,

2−x

x(1−2x)

=

2

x

+

3

(1−2x)

Substituting in equation (1), we get

1−x2

x(1−2x)

=

1

2

+

1

2

(

2

x

+

3

(1−2x)

)
i.e. ∫

1−x2

x(1−2x)

dx = ∫ [

1

2

+

1

2

(

2

x

+

3

(1−2x)

)] dx
= ∫

dx

2

+ ∫

dx

2

+

3

2

‌∫

dx

(1−2x)

=

x

2

+ log |x| +

3

2

×

1

(−2)

log |1 - 2x| + C
=

x

2

+ log |x| -

3

4

log |1 - 2x| + C