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Integral In(x^2-x+2)dx
First I use the intergration by parts, let u=In(x^2-x+2), du=(2x-1)dx/(x^2-x+2), dv=dx, v=x. Then it equals to xIn(x^2-x+2)-integral (x(2x-1))/(x^2-x+2) dx
Then by using long division I get integral (2+(x-4)/(x^2-x+2))dx..but at the end I have no idea how to integral x/(x^2-x+2)...please help with this and thanks a lot.
First I use the intergration by parts, let u=In(x^2-x+2), du=(2x-1)dx/(x^2-x+2), dv=dx, v=x. Then it equals to xIn(x^2-x+2)-integral (x(2x-1))/(x^2-x+2) dx
Then by using long division I get integral (2+(x-4)/(x^2-x+2))dx..but at the end I have no idea how to integral x/(x^2-x+2)...please help with this and thanks a lot.