Q) Integrate [(x^2+1)/(x^2-5x+6)].dx

A)

x^2+1 / x^2-5x+6

= 1 + [ (5x-5) / x^2 -5x + 6 ]

= 1 + [ 5 (x-1+2 -2) / (x-2)(x-3)]

= 1 + 5[ (1 /x-3) + 1/(x-2)(x-3) ]

Notice that 1 in last term can be written as (x-2) - (x-3)

= 1 + 5[(1/x+3) + ( (x-2) - (x-3)) / (x-2)(x-3) ]

= 1 + (10/x-3) - (5/x-2)

integrating we get

x + 10.ln(x-3) - 5.ln(x-2) + c

= x + 5.ln ( (x-3)^2 / (x-2) ) + C

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## Monday, August 11, 2008

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