1. ## Integration partial fraction

Integrate 1/x^2/(x^2+4)

I tried:
A/x + B/(x+1) + (C*x+D)/(x^2+4)

1 = A(x^3+x^2+4*x+4)+B(x^3+4*x)+(C*x+D)*(x^2+x)

1=(A+B+C)*x^3+(A+C+D)*x^2+(4*(A+B)+D)*x+4*A

let x=0..... --> A=1/4
(4*(A+B)+D)=0--> D=-4*B-1 --> D=-1/5
(A+C+D)=0 --> (1/4)+C+-4*B-1=0 --> C=4*B+3/4 --> C= -1/20
(A+B+C)=0 --> (1/4+B+(4*B+3/4) --> B=-1/5

Then I plugged in:
(1/4)/x + (-1/5)/(x+1) + ((-1/20)*x+(-1/20))/(x^2+4)

I still don't know how to get to the solution of the partial fraction step which is supposed to be:
-1/(4*(x^2+4))+1/(4*x^2)

Help me out, I have posted this before and still didn't get it.

2. Hello, LightEight!

Your set-up is wrong . . .

Integrate: . $\int \frac{dx}{x^2(x^2+4)}$

You want: . $\frac{1}{x^2(x^2+4)} \;=\;\frac{A}{x} + {\color{red}\frac{B}{x^2}} + \frac{Cx+D}{x^2+4}$

3. $\int (\frac{1}{x^2(x^2+4)})dx = \frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$

$1=A(x(x^2+4))+B(x^2+4)+(Cx+D)x^2$

$1=Ax^3+4Ax+Bx^2+4B+Cx^3+Dx^2$

$1=(A+C)x^3+(B+D)x^2+4Ax+4B$
$x=0$ $1=4B$ $B=\frac{1}{4}$

$(B+D)=0$ $(\frac{1}{4}+D)=0$ $D=-\frac{1}{4}$

$(A+C)=0$ $A=-C$ I guess I can set $A=0$ and $C=0$

$\frac{0}{x}+\frac{1}{4x^2}-\frac{1}{4x^2+16}$ $-\frac{1}{4x^2+16}+\frac{1}{4x^2}$