<s>[INST] Double integral $ \iint \limits_D \frac{y}{x^2+(y+1)^2}dxdy$, $D$=$\{(x,y): x^2+y^2 \le1 , y\ge0\}$ 
Solve $$ \iint \limits_D  \frac{y}{x^2+(y+1)^2}dxdy \ \ \ \ . . . \  (*)$$
where $D$=$\{$$(x,y): x^2+y^2 \le1 ,  y\ge0       $$\}$

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Here is my attempt.
$$\begin{align}
&(1).\ \ \ (*)=\int_{-1}^1  \int_{0}^{\sqrt{1-x^2}}\frac{y}{x^2+(y+1)^2}dydx \\
&(2).\ \ \ (*)= \int_{0}^1  \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\
&(3). \ \ \int\frac{y+1}{x^2+(y+1)^2}dx = \arctan\left(\frac{x}{y+1}\right) + C  \\
&(4). \ \  \ (*)=\int_{0}^{\pi}  \int_{0}^{1}\frac{r^2sin\theta}{r^2+2rsin\theta+1}drd\theta \\\\
\end{align}$$
I used $(1)$, $(4)$ and $(2)$ with $(3)$, 
but didn't solve yet.
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Did I make a mistake?
Could you give me some advice, please?
How can I solve this integral...
Thank you for your attention to this matter.
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P.S.
Here is result of wolframalpha



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Additionally... I did like this.. maybe useless :-( 
$$\begin{align}
(*) 
&  = \int_{0}^1  \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1  \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{y+1}{x^2+(y+1)^2}dxdy + \int_{0}^1  \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&=\int_{0}^1  \left(\arctan\left(\frac{\sqrt{1-y^2}}{y+1}\right) - \arctan\left(\frac{-\sqrt{1-y^2}}{y+1}\right)\right)dy \\ 
& \ \ \ \ + \int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy  \\\\
&=\int_{0}^1  \left(\arctan\left(\sqrt\frac{1-y}{1+y} \ \right) - \arctan\left(-\sqrt\frac{1-y}{1+y} \ \right)\right)dy \\
& \ \ \ \ +\int_{0}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{x^2+(y+1)^2}dxdy \\\\
&= terrible?!   \\
\end{align}$$
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This picture is for asking to Christian Blatter
(I am really sorry, if I bother you guys for this picture.)

 [/INST] I shall introduce new coordinates (again denoted by $x$, $y$) such that the point $(0,-1)$ becomes the origin, and your vertical axis is my horizontal axis. Your integral then appears as
$$J:=\int_H{x-1\over x^2+y^2}\>{\rm d}(x,y)\ ,$$
where $H$ is the right half of the unit disk with center $(1,0)$. Introducing polar coordinates we obtain
$$J=\int_{-\pi/4}^{\pi/4}\int_{1/\cos\phi}^{2\cos\phi}{r\cos\phi-1\over r^2} r\>dr\ d\phi\ .$$
Here the inner integral evaluates to
$$(2\cos^2\phi-1)-(\log 2+2\log\cos\phi)\ .$$
We therefore get
$$J=1-{\pi\over2}\log 2-4\int_0^{\pi/4}\log\cos\phi\ d\phi=1+{\pi\over2}\log 2-2 \>{\tt Catalan}\doteq0.256862\ ,$$
where ${\tt Catalan}$ is Catalan's constant ($\doteq0.915966$)
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