## anonymous 4 years ago Need help integrating (x+1)/(x^3 + x).

1. slaaibak

${x+1} \over {x(x^2 + 1)}$ Use partial fractions.

2. anonymous

I tried using partial fractions, and got A(x^2 + 1) + (Bx + C)x, but I'm not sure how to get the values of the coefficients.

3. TuringTest

$A(x^2+1)+(Bx+C)x=x+1\to (A+B)x^2+Cx+A$just looking at the coefficients it looks like A=1, C=1, and A+B=0 wich leads to B=-1

4. slaaibak

$x + 1 = A(x^2 + 1) + (Bx + C) x$ $x + 1 = (A+B)x^2 + Cx + A$ Therefore A + B = 0 C=1 A=1 damn turing beat me :/

5. TuringTest

sorry, you got your medal though ;)

6. slaaibak

We will continue this battle in another question ;)

7. TuringTest

indeed! did you get the idea Xylienda?

8. anonymous

:) Thanks guys!

9. anonymous

Yeah. So its more like a trial and error thing, right?

10. TuringTest

Not really in this case when you spell it out and look at the coefficient on the right side you get$(A+B)x^2+Cx+A=x+1$the coefficient of x^2 on the right is zero, so$A+B=0$the coefficient on x is 1, so$C=1$and the constant term is just 1, so$A=1$now that we know A=1 we can go back and solve the first one$A+B=0\to B=-1$so no, it's not trial and error. It is making a system from by setting the coefficients on each side equal. See?

11. anonymous

Oh! I think i really get it now. So you kind of begin by, more or less, matching up the forms on both sides of the equation and compare the coefficients, then work from there.

12. TuringTest

exactly collect like terms on the left and set each coefficient equal to that on the right. Sometimes the systems will be more complex, but here it is pretty easy.

13. anonymous

Thank you! :)