HELP! Integrate (x^2-2)/(x^2-1) dx !?

Question

Integrate (x^2-2)/(x^2-1) dx

(Use partial fractions)

Thanks in Advance

Answer 1

Using partial fractions:
(x² - 2)/(x² -1) = (ax + b)/( x - 1) + c/(x + 1)
Combine the right hand side:
[ax² + ax + bx + b + cx - c]/(x² -1)
This has to be identically equal to (x² - 2)/(x² -1)
So resolving the numerators you get that:
[ax² + ax + bx + b + cx - c] = (x² - 2)
Regroup left hand side:
ax² +(a + b + c)x + (b - c) = (x² - 2)
or
a = 1
a+ b + c = 0
b - c = -2
Solving this system of equations you get
a = 1, b = (-3/2), c = 1/2
So your partial fraction decomposition is:
(x - (3/2)/( x - 1) + (1/2)/(x + 1)
Now integrate
= ∫[(x - (3/2)/( x - 1) + (1/2)/(x + 1)]dx
= ∫x/(x-1)dx + (3/2)ln( x -1) - (1/2)ln(x +1)
= (x -1) + ln(x - 1) + (3/2)ln( x -1) - (1/2)ln(x +1) + C
= (x -1 + C) +(5/2)ln(x - 1) - (1/2)ln(x +1)
= x + (5/2)ln(x - 1) - (1/2)ln(x +1) + C
= x + 5ln[(x - 1)/(x + 1)] + C

Answer 2

(x ² - 2) / (x ² - 1) = 1 - 1 / (x - 1) (x + 1)
Consider 1 / (x - 1) (x + 1):-
1 / (x - 1) (x + 1) = A / (x - 1) + B / (x + 1)
1 = A (x + 1) + B (x - 1)
2A = 1
A = 1 / 2

1 = B( - 2 )
B = - 1 / 2

I = ∫ 1 dx - (1 / 2 ) ∫1 / (x - 1) dx + (1/2)∫1 / (x + 1) dx
I = x - (1/2) log (x - 1) + (1/2) log (x + 1) + C
I = x + (1/2) log [(x + 1) / (x - 1)] + C
I = x + log [(x + 1) / (x - 1)]^(1/2) + log K
I = x + log [ K [(x + 1) / (x - 1) ]^(1/2) ]

Answer 3

(x^2-2)/(x^2-1)=(x^2-1-1)/(x^2-1)
=1- 1/(x^2-1)
=1- 1/(x-1)(x+1)
=1-1/(2(x-1)) +1/(2(x+1))
now u can easily integrate it
the answer is=x +.5(ln(x+1)/ln(x-1)