Question

find integrals of (3x-1)dx/(x-1) (x-2)(x-3).
guyzz please anyone help me​

Answer 1

Solution:-

∫ (3x-1)dx/(x-1)(x-2)(x-3)

(3x-1)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3).........................(1)

(3x-1)/(x-1)(x-2)(x-3) = {A(x-2)(x-3)+B((x-1)(x-3)+C(x-1)(x-2)}/(x-1)(x-2)(x-3)

Here cancel both side (x-1)(x-2)(x-3)

(3x-1) = A(x-2)(x-3) + B((x-1)(x-3)+ C (x-1)(x-2)

⚫ Put x = 1

we get, 2= A(-1)(-2)

A= 2/2= 1

⚫ Put x = 2

we get, 5 = B(1)(-1)

B= -5

⚫ Put x = 3

we get, 8 = C(2)(4)

C= 8/2= 4

We get the value of A,B &C i;e 1,-5 &4

Putting these value in eq (1)

∫ (1/x-1 + (-5)/ x-2 + 4/x-3) dx

= ∫ 1/x-1 dx -5 ∫ 1/x-2 dx + 4 ∫ 1/x-3 dx

( Here we use f'(x)/f(x) dx = log l f(x) l +c)

= log (x-1) -5 log(x-2) +4 log (x-3) +c

Hope it will help you ✌️