Effortlessly Integrate Integral In(x^2-x+2)dx with Proven Techniques

[Song]

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.

 

[chaoseverlasting]

What does In(x^2-x+2)dx mean?
Is it ( 1/(x^2 -x +2) )dx ?

 

[Song]

ln...my bad...

 

[StatusX]

To integrate x/(x^2-x+2), first use substitution to turn this into the integral of C/(x^2-x+2) (ie, write the numerator as 1/2 (2x-1) + 1/2 and take u as the denominator), then factor the numerator as (ax+b)^2+c^2, and finally use the fact that the integral of 1/(x^2+1) is arctan(x).

 

[Song]

sorry, I don't get how "To integrate x/(x^2-x+2), first use substitution to turn this into the integral of C/(x^2-x+2) (ie, write the numerator as 1/2 (2x-1) + 1/2 and take u as the denominator)" works...

Do you mean let x=(1/2)(2x-1)+1/2, then it will be 1/2 integral 2x/(x^2-x+2) dx, then use U substitution u=x^2 then it becomes integral (u-sqrtu+2)^(-1) du?

I'm lost...

 

[StatusX]

Yes, sorry, that wasn't very clear. I meant, write:

\int \frac{x}{x^2-x+2} dx = \int \frac{1/2(2x-1) +1/2}{x^2-x+2} dx=\frac{1}{2} \int \frac{2x-1}{x^2-x+2} dx + \frac{1}{2} \int \frac{1}{x^2-x+2} dx

The first term can be integrated by substitution, so you're left with the second term to integrate. That's what I meant by "turn it into the integral of C/(x^2-x+2)". Do you understand what to do from here?

 

[Song]

yes. Thanks so much!