Talk:Calculus/Integration techniques/Partial Fraction Decomposition

How do you integrate something like this: 2/ (x+2)(x^2+3). Also when it is in the A/x + B/x form, can the denominator x have a power greater than one?


 * To integrate something like $$\frac{3}{(x+2)(x^2+3)}$$ you try to write the fraction as
 * $$\frac{3}{(x+2)(x^2+3)}=\frac{A}{(x+2)}+\frac{Bx+C}{x^2+3}$$


 * Adding the terms together on the right hand side we get:
 * "$$\frac{3}{(x+2)(x^2+3)}=\frac{A(x^2+3)+(Bx+C)(x+2)}{(x+2)(x^2+3)}=\frac{(A+B)x^2+(2B+C)x+(3A+2C)}{(x+2)(x^2+3)}$$


 * From looking at the $$x^2$$ term we see we need $$A+B=0$$, or $$A=-B$$. By looking at the $$x$$ term we need $$2B+C=0$$, or $$C=2A$$.  From the constant term we see we need $$3A+2C=3$$, or $$7A=3$$.  Thus $$A=3/7$$, $$B=-3/7$$ and $$C=6/7$$.


 * You can get powers bigger than one in the denominator if you start off with repeated roots. For example to do a problem like $$\frac{1}{x^3(x+2)}$$, you'll need terms that look like $$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+2}$$. Thenub314 (talk) 03:43, 17 May 2011 (UTC)