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