Find the indefinite integral using integration by partial fractions: S2x/((x^2)-4 dxThis is what I have figured: 2x/[(x-2)(x+2)] = A/(x-2) + B/(x+2) 2x=A(x+2) + B(x-2) let x=-2 2(-2) = A(-2+2)...
Find the indefinite integral using integration by partial fractions: S2x/((x^2)-4 dx
This is what I have figured:
2x/[(x-2)(x+2)] = A/(x-2) + B/(x+2)
2x=A(x+2) + B(x-2) let x=-2
2(-2) = A(-2+2) +B(-2-2)
-4=-4B So B=1 then let x=2
2(2) = A(2+2) + B(2-2)
4=4A So A=1
so S(1/(x-2) +1/(x+2))dx
S(1/x-2)dx + S(1/x+2)dx
ln|x-2|+ln|x+2|+C
please let me know if I am on the right track with this one.
Thanks
1 Answer
`(2x)/(x^2-4) = (2x)/((x-2)(x+2))`
Using partial fractions;
`(2x)/((x-2)(x+2)) = A/(x-2)+B(x+2)`
`2x = A(x+2)+B(x-2)`
`2x = x(A+B)+2(A-B)`
When you consider x; in the left side the component of x (the number with x term) is 2 and the right side it is (A+B).
So;
`2 = A+B -----(1)`
When you consider constant term ; in the left side the component of constant is 0 and the right side it is 2(A-B).
`0 = 2(A-B) `
`0 = A-B ----(2)`
Once you solve (1) and (2);
`A = 1`
`B = 1`
So we can write;
`(2x)/((x-2)(x+2)) = 1/(x-2)+1(x+2)`
`int (2x)/((x-2)(x+2))dx`
`= int1/(x-2)+1(x+2) dx`
`= int 1/(x-2)dx+int1/(x+2)dx`
`= ln(x-2)+ln(x+2)+C` C is a constant.
`= ln((x-2)(x+2))+C`
So
`int (2x)/((x-2)(x+2))dx = ln((x-2)(x+2))+C`
Note:
Your method is correct. This is another way that may be easy to use with complex situations.