We have a dipole and I need to find $V(x)$ for $x<-d$

First I calculate the potential at some point $A=(x,0)$ relative to $P = (-\infty, 0) $ of the $-Q$ charge as follows $$\int_A^P\vec {E} \cdot \vec {dl}=\int_A^P\frac{Q\vec i}{4\pi\epsilon_0(-d-x)^2}dx(-\vec i)=-\frac{Q}{4\pi\epsilon_0}\int_A^P\frac{dx}{(d+x)^2}=\frac{Q}{4\pi\epsilon_0}\frac{1}{d+x}\biggr |_A^P=$$ $$= -\frac{Q}{4\pi\epsilon_0}\frac{1}{d+x}= V(x)$$

And then I repeat the same procedure for the $+Q$ charge and then add the potentials up. I get the right functional form but my sign is reversed so I get that the potential for $x < -d$ is actually positive when it's not. However, I just can't find what am I doing wrong.

The equation that I get for the potential by the $+Q$ charge is $$V_{+Q}(x)=-\frac{Q}{4\pi\epsilon_0}\frac{1}{d-x}$$

The right answer should be $$V(x) = -\frac{2Qd}{4\pi\epsilon_0(x^2-d^2)}$$ but I repeatedly fail to obtain the minus sign in front.


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