Question

int 6x/(x^3-8)

Answer 1

x^3 - 8 = x^3 - 2^3
....
use partial fraction method

Answer 2

i get it down to int 1/(x-2)+int (-x+2)/s^2+2x+4
i know the first part ln (x-2) but the second is losing me

Answer 3

okay int (2-x)/(x+1)^2-3

Answer 4

(-x+2) = - (x + 1) + 3
and try completing squares ...

Answer 5

For the second part, you can split the (2-x)/(x^2+2x+4) into 3/(x^2+2x+4) and -(1/2)*(2x+2)/(x^2+2x+4), the second part of which is easily integrable and the first part of which you can do complete the square on and then substitute.

Answer 6

how do you split the 2-x into 3 and 1/2 * 2x+2 ?

Answer 7

i see the sub once you get to that point.

Answer 8

Don't forget the negative. -1/2*(2x+2)=-x-1, add 3 and you get 2-x.

Answer 9

If you mean how do you get that, you just do whatever manipulation you need to get the 2x+2 term that you need to integrate

Answer 10

i am just not see how to split (2-x) into 3 + (-1/2)*(2x+2)

Answer 11

http://www.wolframalpha.com/input/?i=simplify+3+%2B+%28-1%2F2%29*%282x%2B2%29
that a common trick you will need often

Answer 12

You need a 2x+2 term to integrate as log(x^2+2x+4), right? So you simply do algebraic manipulation to get a 2x+2 term. It happens that the way to do that is by splitting it into 3/(x^2+2x+4) and (-1/2)(2x+2)/(x^2+2x+4).

Answer 13

okay got it thanks