`(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.
We’ll help your grades soar
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support
Already a member? Log in here.
Are you a teacher? Sign up now