Definite integration. $$\int _0^1 \arctan(x^2-x+1)\,dx$$
ATTEMPT:
$\int \arctan(x^2-x+1)\,dx$
Let
$\arctan(x^2-x+1)=u$ and $dx=dv$: 
$$du=\frac{2x−1}{(x^2−x+1)^2+1}dx= \frac{2x−1}{(x^2+1)(x^2-2x+2)}$$ and $v=x$.
Now from Integration by parts $I=uv-$$\int vdu.$
$$I=\arctan(x^2-x+1)x-\int \frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}.$$
Now using partial fractions
$$
\int\frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}=\int \frac{x}{(x^2+1)}-\frac{x}{x^2-2x+2} $$
Integrating term by term and finally arranging the result:
$$I=x\arctan(x^2−x+1)+\frac{1}{2}\ln(∣x^2+1∣)−\frac{1}{2}\ln(∣x^2−2x+2∣)−\arctan(x−1).$$
Now substituting the limit i got $I=\ln2.$
Is there any other way to approach this problem like by using the properties of definite integration as this method is long and tedious but though it works!!
 A: We have $$\int_0^1\arctan(x^2-x+1)dx =\int_0^1\left (\arctan(\frac1x)+\arctan(x-1) \right)dx=   
\\\\ \left [x\cot^{-1}(x)+\frac12 \log(1+x^2)\right] \Biggr|_0^1+\left[(1-x)\tan^{-1}(1-x)-\frac12 \log(1+(x-1)^2)\right] \Biggr|_0^1=\log2$$
A: This response is similar in spirit to @nospoon's answer, but I think it is perhaps more elementary because it doesn't require the anti-derivative of arctangent.
$$\begin{align}
I
&=\int_{0}^{1}\arctan{\left(x^2-x+1\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}--\arctan{\left(\frac{1}{x^2-x+1}\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-\arctan{\left(x\right)}-\arctan{\left(1-x\right)}\right]\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\arctan{\left(1-x\right)}\,\mathrm{d}x\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x-2\int_{0}^{1}\arctan{\left(x\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{\pi}{2}-2\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\operatorname{arccot}{\left(x\right)}-\arctan{\left(x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{2x}{1+x^2}\,\mathrm{d}x\\
&=\ln{\left(1+x^2\right)}\bigg{|}_{0}^{1}\\
&=\ln{(2)}.\\
\end{align}$$
