How to solve dis partial fraction?

Question

Qns: 3/ x(3x - 1)^2
Ans: 3/x + 9/ (3x -1) + 9/ (3x -1)^2

can help to solve...plz help to write every step down to solve dis qns

Answer 1

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

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

Let x = 1 / 3:-
3 = (1/3) C
C = 9

Let x = 0:-
3 = A


Let x = 2
3 = 25A + 10B + 18
3 = 75 + 10B + 18
10B = - 90
B = - 9

Answer is:-
3 / x - 9 / (3x - 1) + 9 / (3x - 1)²

Answer 2

=A/x + B/(3x-1) + C/(3x-1)^2
Take LCM on RHS
hence 3= A(3x-1)^2 + B(3x-1)x + Cx
(As the denominators wil get canceld)

Let's call dat equation as 1.

Now put x=0 in eq.1.
Thus we'll get the value of A.
Now put x=1/3 in eq1.
Thus, we'll get value of C.
Now,put the values of A and C,aswell as x=1 in eq1.
Thus we'll get the value of B...
Done!!!!

Answer 3

remedy crucial a million/(x^3-^2) dx enable, A/x + B/(x^2) + C/(x-5) = a million/(x^3-^2) = a million/x^2(x-5) Multiply each and every part by x^2(x -5), we get, A(x^2)(x-5) + B(x)(x-5) + C(x)(x^2) = a million Equating the coefficient, we get, A = -a million/25, B = a million/5, C = a million/25 So, crucial a million/(x^3-^2) dx = crucial[-a million/25x +a million/x^2 +a million/25(x-5)].dx = (-a million/25)ln(x) + a million/5x + (a million/25)ln(x-5) +C >============< answer

Answer 4

3/ x(3x - 1)^2
=3/(3x-1)[1/(x(3x - 1))]
=3/(3x-1)[3/(3x-1)-1/x]
=9/(3x-1)^2 - 3/ x(3x - 1)
=9/(3x-1)^2 - 3[3/(3x-1)-1/x]
=9/(3x-1)^2 +3 /x +9/(3x-1)