Hello all!
Using (3/x-2) - (2/x-1) calculate the area enclosed between the curve
y= (x+1)/(x-2)(x-1) and the x axis, between x = 3 and x = 5
I would like to confirm whether the work I have done is correct or not
and if anyone could give me some guidance
∫3/x-2 dx - ∫2/x-1
3∫1/x-2 dx + 2∫1/x-1 dx
5
[3ln|x-2| + 2ln|x-1|]
3
(3ln|5-2| + 2ln|5-1|) - (3ln|3-2| + 2ln|3-1|)
= 4.682 units squared