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.)

 A: 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 ($\doteq0.915966$)
A: The integrand function $\frac{y}{x^2+(y+1)^2}$ suggest to put
$$\left\{
\begin{align}
x&=r\cos\theta\\
y+1&=r\sin\theta
\end{align}\right.
$$
so that $x^2+(y+1)^2=r^2$ and the Jacobian is $r$.
From $y\ge 0$ we have $y=r\sin\theta-1\ge 0$ that is $r\ge\frac{1}{\sin\theta}$ and from $x^2+y^2\le 1$ we have $$x^2+y^2=r^2\cos^2\theta+(r\sin\theta-1)^2=r^2-2r\sin\theta+1\le 1$$ and then $r(r-2\sin\theta)\le0$ so that $r\le 2\sin\theta$.
So we have
$$\boxed{
r_{\min}=\frac{1}{\sin\theta}\le r\le 2\sin\theta=r_{\max}}
$$
For $y=0$ (i.e. $r\sin\theta =1$) we have $-1\le x\le 1$, that is $-1\le r\cos\theta\le 1$ and then $-1\le\tan\theta\le 1$; thus
$$\boxed{
\theta_{\min}=\frac{\pi}{4}\le \theta\le \frac{3\pi}{4}=\theta_{\max}}
$$
or $\frac{-\pi}{4}\le \theta\le \frac{+\pi}{4}$ if you prefer.
The figure help to show all we have done.

So the integrand in polar coordinates becomes $f(r,\theta)=\frac{r\sin\theta-1}{r^2}$ and the integral becomes
$$
\mathcal{I}=\int_{\theta_{\min}}^{\theta_{\max}}\int_{r_{\min}}^{r_{\max}}
f(r,\theta)r\,\mathrm d r\,\mathrm d\theta=
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{\frac{1}{\sin\theta}}^{2\sin\theta}\left(\sin\theta-\frac{1}{r}\right) \mathrm d r\,\mathrm d\theta
$$
The integral in $r$ is easy to evaluate
$$\begin{align}
\int_{\frac{1}{\sin\theta}}^{2\sin\theta}\left(\sin\theta-\frac{1}{r}\right) \mathrm d r
&=
\left[\sin\theta\, r-\log r\right]_{\frac{1}{\sin\theta}}^{2\sin\theta}\\
&=\sin\theta\left[2\sin\theta-\tfrac{1}{\sin\theta}\right]-\left[\log(2\sin\theta)-\log\left(\tfrac{1}{\sin\theta}\right)\right]\\
&=-\cos(2\theta)-\log\left(2\sin^2\theta\right)
\end{align}
$$
Then the integral in $\theta$ is
$$\begin{align}
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left[-\cos(2\theta)-\log\left(2\sin^2\theta\right)\right]\mathrm d \theta
&=
\left[-\frac{1}{2}\sin(2\theta)\, \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(2\sin^2\theta\right)\mathrm d \theta=1+J
\end{align}
$$
where
$$
J=-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(2\right)\mathrm d \theta-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(\sin^2\theta\right)\mathrm d \theta=
-\frac{\pi}{2}\log 2-2C+\pi\log 2=\frac{\pi}{2}\log 2-2C
$$
observig that 
$$\begin{align}
-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \log\left(\sin^2\theta\right)\mathrm d \theta &=
-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\cos^2\theta\right)\mathrm d \theta=-2\int_{0}^{\frac{\pi}{4}} \log\left(\cos^2\theta\right)\mathrm d \theta\\
&=-4\int_{0}^{\frac{\pi}{4}} \log\left(\cos\theta\right)\mathrm d \theta=-4\left(\frac{C}{2}-\frac{\pi}{4}\log 2\right)\\
&=-2C+\pi\log 2
\end{align}
$$
where $C$ is the Catalan's constant (see for exaple here).
Finally we have
$$\large\color{blue}{
\mathcal I=1+\frac{\pi}{2}\log 2-2C}
$$
