∫

ex

√5−4ex−e2x

dx
Let ex = t , ex dx = dt
Now integral I becomes,
I = ∫

dt

√5−4t−t2

⇒ I = ∫

dt

√5+4−4−4t−t2

⇒ I = ∫

dt

√9−(4+4t+t2

⇒ I = ∫

dt

√32−(t+2)2

⇒ I = sin−1

(t+2)

3

+ C
⇒ I = sin−1

(ex+2)

3

+ C
OR

(x−4)ex

(x−2)3

dx
I = ∫ex(

x−2

(x−2)3

−

2

(x−2)3

) dx
I = ∫ ex(

1

(x−2)2

−

2

(x−2)3

) dx
Thus the given integral is of the form,
I = ∫ ex |f (x) + f' (x)| dx where , f (x) =

1

(x−2)2

; f' (x) =

−2

(x−2)3

I = ∫

ex

(x−2)2

dx - ∫

2ex

(x−2)3

dx
=

ex

(x−2)2

- ∫

ex(−2)

(x−2)3

dx - ∫

2ex

(x−2)3

dx + C
So, I =

ex

(x−2)2

+ C