?

Log in

Solved Integrals (1) - Mathematicus Ridiculus [entries|archive|friends|userinfo]
Mathematicus Ridiculus

[ userinfo | livejournal userinfo ]
[ archive | journal archive ]

Solved Integrals (1) [Aug. 18th, 2006|10:03 pm]
Mathematicus Ridiculus
mathphiles
[zheyyy]
[Tags|, ]
[Current Location |Some random place.]
[mood |stressedstressed]
[music |Keane - Somewhere only we know]

Integrate ⌠(1/(x^2-1)) dx
First, try to split this expression up into partial fractions.

Partial Fractions
Let 1/(x2-1) = A/(x-1) + B/(x+1)
= (Ax+A+Bx-B)/( x^2-1)
Thus,
Ax+A+Bx-B = 1
(A+B)x + (A-B) = 0x + 1
A+B = 0 ---------------------1
A-B = 1 ---------------------2
Solving simultaneous equations,
A = ½
B = -½

Therefore,
1/(x^2-1) = 1/(2x-2) – 1/(2x+2)

Which makes integration really easy.

⌠(1/(x^2-1)) dx
= ⌠(1/(2x-2) – 1/(2x+2)) dx
= ⌠(1/(2x-2))dx - ⌠(1/(2x+2)) dx
= ½ (ln(2x-2) – ln(2x+2)) + C
linkReply