How do I find the following definite integral? Find the value of the following definite integral:
$$\int_0^{1}\frac{\arctan(t)}{1+t}dt$$
I tried using integration by parts, but it gives another complicated integral, so possibly the antiderivative does not exist. What other methods can I apply?
 A: Let us write $$I=\int_0^1\dfrac{\arctan t}{1+t}dt$$
The change of variables  $t=\dfrac{1-x}{1+x}$ shows that
$$\eqalign{I&=\int_0^1\arctan\left(\frac{1-x}{1+x}\right)\frac{dx}{1+x}\\
&=\int_0^1\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{1+x}\\
&=\frac{\pi}{4}\int_0^1\frac{dx}{1+x}-\int_0^1\frac{\arctan x}{1+x}dx\\
&=\frac{\pi}{4}\ln 2-I
}$$
Thus $I=\dfrac{\pi}{8}\ln 2$.$\qquad \square$
A: It is not as easy an integral as it first appears to be. I am cutting/pasting solutions in four cases obtained by Mathematica Version 8. ...including the upper limit 1 you changed. 
First case result is real, even if it involves $i$ but fails to annul $i$ in the final expression.
Second case is what you asked for,  evaluated numerically.
Third one actually involves Arctan, Log, and Catalan constant. 
Fourth case with upper limit 1 has an easily obtainable result, but connection to simplification due the upper limit 1  is surprising...
$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),u] $
$
1/32 (-5 I[\text{Pi}]{}^{\wedge}2+8 I[\text{Pi}] \text{ArcTan}[u]-32 I \text{ArcTan}$ 
$[u]{}^{\wedge}2-24[\text{Pi}] \text{Log}[2]+16[\text{Pi}] \text{Log}$
$[1+E{}^{\wedge}(-2 I \text{ArcTan}[u])] $
$-32 \text{ArcTan}[u] \text{Log} [1+E{}^{\wedge}(-2 I \text{ArcTan}[u])]+ $
$ 8[\text{Pi}] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])] $ 
$ +32 \text{ArcTan}[u] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])]+ $
$8[\text{Pi}] \text{Log}[1+u{}^{\wedge}2]-8[\text{Pi}] \text{Log}$ $[\text{Sin}$
$[[\text{Pi}]/4+  \text{ArcTan}[u]]]-16 I \text{PolyLog}[2,-E{}^{\wedge}(-2 I $ 
$\text{ArcTan}[u])]-16 I \text{PolyLog}[2,I E{}^{\wedge}(2 I \text{ArcTan}[u])]) $
$ \text{NIntegrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4.\}] $
0.189728
$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4\}] $
$ 1/96 (-48 \text{Catalan}-47 I[\text{Pi}]{}^{\wedge}2+[\text{Pi}] (24 I$
$ \text{ArcTan}[[\text{Pi}]/4]-2$ 
$\text{Log}[4096]+24 (\text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}$
$[[\text{Pi}]/4])]+2 \text{Log}[I/(-4 I+[\text{Pi}])]+\text{Log}[16+[\text{Pi}]$
${}^{\wedge}2]+\text{Log}[ \text{Csc}[[\text{Pi}]/4+ $ 
$ \text{ArcTan} [[\text{Pi}]/4]]]))-48 I (2 \text{ArcTan}[[\text{Pi}]/4] $
$ {}^{\wedge}2+2I \text{ArcTan}[[\text{Pi}]/4] (\text{Log}[-((1+I)/(4 $
$ I+[\text{Pi}]))]-\text{Log}[(8 I)/((-4 I+[\text{Pi}]) $ 
$ (4+[\text{Pi}]))])+\text{PolyLog}[2,-((4+I[\text{Pi}])/(4 $
$ I+[\text{Pi}]))]+\text{PolyLog}[2,(4 I+[\text{Pi}])/(-4 I+[\text{Pi}])])) $
$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,1\}] $
$ log (2) \pi/8  $
The given function is regular, is a product of two functions one increasing, other decreasing, so has a maximum, ( at t = 1.22913).
