How do you find $int(x-1) / ( 2x^2 - 3x -3) dx$ using partial fractions?

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Answer 1 · 2016-06-03T18:24:37

$int (x-1)/(2x^2-3x-3) d x=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C$

$int (x-1)/(2x^2-3x-3) d x=?$

$2x^2-3x-3=u" ; "d u=(4x-3)*d x$

$=1/4int (4(x-1))/(2x^2-3x-3) d x$

$=1/4[ int (4x-3)/(2x^2-3x-3) d x-1/(2x^2-3x-3) d x]$

$=1/4[int (d u)/u-(d x)/(2x^2-3x-3)]$

$Delta=sqrt(b^2-4*a*c)$

$Delta=sqrt((-3)^2+4*2*3)=sqrt(33)$

$=1/4 l n|u|-1/4int (d x)/(2x^2-3x-3)$

$=1/4 l n| 2x^2-3x-3|-1/4int (d x)/(2x^2-3x-3)$

$=1/4 l n|2x^2-3x-3|-1/4 1/Delta l n|(2ax+b-Delta)/(2ax+b+Delta)|+C$

$2ax=2*2*x=4x" ;"b=-3" ; "Delta =sqrt33$

$=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C$

$int (x-1)/(2x^2-3x-3) d x=1/4 l n|2x^2-3x-3|-1/4 1/sqrt33 l n|(4x-3-sqrt33)/(4x-3+sqrt 33)|+C$

How do you find int(x-1) / ( 2x^2 - 3x -3) dxint(x-1) / ( 2x^2 - 3x -3) dx using partial fractions?