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

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

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

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

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.

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