1. ## [SOLVED] Antiderivative

$\int{\frac{x-1}{(x+1)^2(x^2+3)}}dx = \int{\left(\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C x + D}{x^2+3}\right)dx}$

$A(x+1)^2(x^2+3) + B(x+1)(x^2+3) + (Cx + D)(x+1)^3 = x-1 \implies$

$(A+C)x^4 + (2A+B+3C+D)x^3 + (4A+B+3C+3D)x^2 +
(6A+3B+C+3D)x +$
$(3A+3B+D) = x-1$

Good so far? I fail to properly solve the equation system every time. I could really need another pair of eyes to look at what I've done so far, because this is driving me crazy.

$x^4: A+C=0$
$x^3: 2A+B+3C+D=0$
$x^2: 4A+B+3C+3D =0$
$x^1: 6A+3B+C+3D=1$
$x^0: 3A+3B+D=-1$

2. Seeing the answer may help see how to do it.

3. Originally Posted by Spec
$\int{\frac{x-1}{(x+1)^2(x^2+3)}}dx = \int{\left(\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C x + D}{x^2+3}\right)dx}$

$A(x+1)^2(x^2+3) + B(x+1)(x^2+3) + (Cx + D)(x+1)^3 = x-1 \implies$

Mr F says: No. You've made a lot of mistakes. It should be:

${\color{red}A(x+1)(x^2+3) + B(x^2+3) + (Cx + D)(x+1)^2 = x-1}$.

The simplest way is to then substitute convenient values of x into the above. For example, x = -1 will let you get B very easily.

[snip]
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