Question

Need help integrating (x+1)/(x^3 + x).

Answer 1

\[{x+1} \over {x(x^2 + 1)}\]

Use partial fractions.

Answer 2

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.

Answer 3

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

Answer 4

\[x + 1 = A(x^2 + 1) + (Bx + C) x\]

\[x + 1 = (A+B)x^2 + Cx + A\]

Therefore A + B = 0

C=1

A=1

damn turing beat me :/

Answer 5

sorry, you got your medal though ;)

Answer 6

We will continue this battle in another question ;)

Answer 7

indeed!
did you get the idea Xylienda?

Answer 8

:) Thanks guys!

Answer 9

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

Answer 10

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?

Answer 11

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.

Answer 12

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.

Answer 13

Thank you! :)