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I have this integral: $$ \int \frac{1}{(1+x^2)(1+(z-x)^2)} {\rm d}x $$ and I want to perform partial fraction decomposition in this form $$ \int \left( \frac{Ax + B}{1+x^2} + \frac{Cx + D }{1+(z-x)^2} \right) {\rm d}x $$ but I can't get the right coefficients A, B, C, D. Is this the right way to do it?

Edit: My mistake was in expanding the brackets $$ (Ax+B)(1+(z-x)^2)+(Cx+D)(1+x^2)=1 $$ particularly, instead of $Axz^2$ I arrived to $Az^2$ and then I was trying to solve completely dfferent system of equations.

Raffer Flameline
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This is correct $$\frac{1}{(1+x^2)(1+(z-x)^2)}=\frac{Ax + B}{1+x^2} + \frac{Cx + D }{1+(z-x)^2}$$ Reduce to the common denominator and expand the rhs; you then have $$1=\left(B z^2+B+D\right)+x \left(A z^2+A-2 B z+C\right)+x^2 (-2 A z+B+D)+x^3 (A+C)$$ So, you hve to solve for $(A,B,C,D)$ $$B z^2+B+D=1 \tag 1$$ $$A z^2+A-2 B z+C=0 \tag 2$$ $$-2 A z+B+D=0 \tag 3$$ $$A+C=0 \tag 4$$ which is not difficult.

Claude Leibovici
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