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[QUOTE="HallsofIvy, post: 4493512, member: 637751"] So the two numerators are equal: [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex] Taking x= 2 gives 8- 18+ 8= 8C which gives C= -1/4, as you have below, not 0. Did you forget the left side of the equation? If, in [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex], you take x= 0 (just because it is an easy number) you get [itex]8= (0+ B)(-2)+ C(4)[/itex] or [itex]-2B= -8- (-1/4)(4)= -9[/itex] so that C= -9/2, again as you say. Finally, taking x= 1 (again just because it is an easy number) gives [itex]2- 9+ 8= (A+ B)(-1)+ C(5)[/itex] or [itex]1= -A+ 9/2- 5/4[/itex] so that [itex]A= 9/2- 1- 5/4= 18/4- 4/4- 5/4= 9/4[/itex], again, exactly what you have. Great! Now all you need to do is the integral: [tex]\int dx+ \dfrac{9}{4}\int \dfrac{xdx}{x^2+ 4}- \dfrac{9}{2}\int \dfrac{dx}{x^2+ 4}- \dfrac{1}{4}\int \dfrac{dx}{x- 2}[/tex] [/QUOTE]
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