integrate (x^2-x+6)/(x^3+3x) dx
I get stuck when im trying to find what a b and c should be.
integrate (x^2-x+6)/(x^3+3x) dx
I get stuck when im trying to find what a b and c should be.
I don't know what a, b or c are and it looks like you've tried to include an image; could you edit the post please so that I can help?
What "a", "b", and "c"? There is only "x" in your integrand.Originally Posted by summergrl
Please reply with clarification, including a clear listing of all of your work and reasoning so far.
Thank you.
Eliz.
sorry...the image was there by mistake.
I am pretty sure I need to use integration by parts and we use A, B, and C.
So so far i seperated it into
A/X + (Bx+C)/(x^2+3)
I think you mean partial fractions, not integration by parts.
[tex]\L \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3}[/tex]
[tex]\L \frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{A(x^2+3)}{x(x^2+3)} + \frac{x(Bx + C)}{x(x^2 + 3)}[/tex]
[tex]\L x^2 - x + 6 = A(x^2 + 3) + x(Bx + C)[/tex]
[tex]\L x^2 - x + 6 = (A + B)x^2 + Cx + 3A[/tex]
equate coefficients ...
[tex]\L A + B = 1[/tex]
[tex]\L C = -1[/tex]
[tex]\L 3A = 6[/tex]
can you finish?
oops! i keep on mixing those two names up.
how do you know A+b=1?
[tex]\L x^2 - x + 6 = (A+B)x^2 + Cx + 3A[/tex]
look at the left side of the equation ... what is the coefficient of x<sup>2</sup>?
look at the right side of the equation ... what is the coefficient of x<sup>2</sup>?
now set the coeffcients equal.
If you know about complex numbers you can do it much faster...
... and if only pigs could fly.Originally Posted by Count Iblis
okay thanks! i understand that part now...so I got B=-1, C=-1, and A =2.
then i get the integration of
(2/x) - (x/x^2+3) - (1/x^2+3)
and i get 2lnx - 1/2ln(x^2+3) but then i don't know how to do the third part. Does it have anything to do with arctan?
p.s. whats this talk about pigs?? hehe
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